The mechanical energy of air per unit mass is 50 J/kg.
The power generation potential of a wind turbine with 85-m-diameter blades at that location is approximately 147.8 kW.
The mechanical energy of air per unit mass can be calculated using the formula:
Mechanical energy per unit mass = (1/2) * v^2
where v is the velocity of the air.
Given that the wind velocity is 10 m/s, we can substitute this value into the formula:
Mechanical energy per unit mass = (1/2) * (10 m/s)^2
Mechanical energy per unit mass = (1/2) * 100 J/kg
Mechanical energy per unit mass = 50 J/kg
Power = (1/2) * ρ * A * v^3
where ρ is the air density, A is the area swept by the blades, and v is the velocity of the wind.
Given that the air density (ρ) is 1.25 kg/m³ and the diameter (d) of the blades is 85 m, we can calculate the area swept by the blades (A):
A = π * (d/2)^2
A = π * (85 m/2)^2
A = 5669.91 m²
Power = (1/2) * (1.25 kg/m³) * (5669.91 m²) * (10 m/s)^3
Power ≈ 147,810 W
Converting the power to kilowatts:
Power ≈ 147.8 kW
The mechanical energy of air per unit mass is 50 J/kg. The power generation potential of a wind turbine with 85-m-diameter blades at that location is approximately 147.8 kW.
These values are obtained by calculating the mechanical energy per unit mass based on the wind velocity and the power generated by the wind turbine using the air density, blade diameter, and wind velocity.
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The position of an object is time is described by this equation x=414-71° +21 - 81 +11 a Write an equation of the objects velocity as a function of time. b Write an equation of the objects acceleration as a function of time.
(a) The equation for the object's velocity as a function of time is v(t) = -71t + 21. (b) Since the given position equation does not include a term for acceleration, the acceleration is constant and its equation is a(t) = 0.
(a) The position equation x(t) = 414 - 71t + 21 - 81 + 11 describes the object's position as a function of time. To find the equation of the object's velocity, we differentiate the position equation with respect to time.
The constant term 414 and the other constants do not affect the differentiation, so they disappear. The derivative of -71t + 21 - 81 + 11 with respect to t is -71, which represents the velocity of the object. Therefore, the equation of the object's velocity as a function of time is v(t) = -71t + 21.
(b) To find the equation of the object's acceleration, we differentiate the velocity equation v(t) = -71t + 21 with respect to time. The derivative of -71t with respect to t is -71, which represents the constant acceleration of the object.
Since there are no other terms involving t in the velocity equation, the acceleration is constant and does not vary with time. Therefore, the equation of the object's acceleration as a function of time is a(t) = 0, indicating that the acceleration is zero or there is no acceleration present.
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Light is travelling from medium A(refractive index
1.4) to medium B ( refractive index 1.5). If the incident angle is
44.3⁰, what would be the refracted angle in medium B? Express
answer in degrees
The refracted angle in medium B, when light travels from medium A to medium B, is approximately 41.3 degrees.
To find the refracted angle in medium B when light travels from medium A to medium B, we can use Snell's Law. Snell's Law states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the refractive indices (n₁ and n₂) of the two mediums:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In this case, the incident angle (θ₁) is given as 44.3 degrees, and the refractive indices of medium A and medium B are 1.4 and 1.5, respectively.
Let's plug in the values and solve for the refracted angle (θ₂):
1.4 * sin(44.3°) = 1.5 * sin(θ₂)
θ₂ = arcsin((1.4 * sin(44.3°)) / 1.5)
Evaluating the equation, we find that the refracted angle in medium B is approximately 41.3 degrees. Therefore, the refracted angle in medium B is 41.3° (rounded to one decimal place).
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Physical Science
Based on the data given in the Periodic Table of Elements in your classroom, calculate the formula mass for H2SO4 (sulfuric acid).
Formula mass of sulfuric acid (H2SO4)The chemical formula for sulfuric acid is H2SO4. The formula mass is the sum of the masses of the atoms in the molecule.
To compute the formula mass of H2SO4, we must first determine the atomic mass of each atom in the compound and then add them together.
Atomic masses for H, S, and O are 1.008, 32.06, and 16.00, respectively.
Atomic mass of H2SO4 is equal to (2 x 1.008) + 32.06 + (4 x 16.00)
= 98.08 g/mol
Therefore, the formula mass of sulfuric acid (H2SO4) is 98.08 g/mol.
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A uniform 10kg, 2m, horizontal tree branch is attached to a tree on the left side. At the far right end, a vine is wrapped around the branch and is pulling on it at an angle of 20° above the positive z-axis. Two birds are perched on the branch: a .02kg bird. Im from the left end of the branch and a .05kg bird .3m from the right end of the branch. The birds and branch are completely motionless. (a) What's the tension in the vine? (b) What are the z and y components of the support force exerted by the tree on the branch?
(a) The tension in the vine is equal to the weight of the branch plus the weights of the birds on the branch. (b) The z-component of the support force exerted by the tree on the branch is equal to the tension in the vine, while the y-component is the sum of the weights of the branch and the birds.
(a) The tension in the vine can be determined by considering the equilibrium of forces acting on the branch. Since the birds and the branch are motionless, the net force in the vertical direction must be zero. First, let's find the vertical components of the weights of the birds:
Weight of the first bird = m1 * g = 0.02 kg * 9.8 m/s^2 = 0.196 N
Weight of the second bird = m2 * g = 0.05 kg * 9.8 m/s^2 = 0.49 N
The total vertical force acting on the branch is the sum of the weights of the birds and the tension in the vine:
Total vertical force = Weight of first bird + Weight of second bird + Tension in the vine
Since the branch is in equilibrium, the total vertical force must be zero:
0.196 N + 0.49 N + Tension in the vine = 0
Solving for the tension in the vine:
Tension in the vine = -(0.196 N + 0.49 N) = -0.686 N
Therefore, the tension in the vine is approximately 0.686 N.
(b) The support force exerted by the tree on the branch has both z and y components.
The z-component of the support force can be determined by considering the equilibrium of torques about the left end of the branch. Since the branch and birds are motionless, the net torque about the left end must be zero.
The torque due to the tension in the vine is given by:Torque due to tension = Tension in the vine * Distance from the left end of the branch to the point of application of tension
Since the branch is in equilibrium, the torque due to the tension must be balanced by the torque due to the support force exerted by the tree. Therefore:
Torque due to support force = -Torque due to tension
The y-component of the support force can be found by considering the vertical equilibrium of forces. Since the branch and birds are motionless, the net force in the vertical direction must be zero.
The z and y components of the support force exerted by the tree on the branch can be determined by solving these equations simultaneously.
Given the values and distances provided, the specific magnitudes of the z and y components of the support force cannot be determined without additional information or equations of equilibrium.
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Concave Converging Ray Diagrams 1. An object is located 14 cm in front of a concave mirror. If the focal length is 3 cm, locate the object and draw the ray diagram for the resulting image. Object Type (Real or Virtual): Orientation (Upright or Inverted): Location (front or behind): Size (same, larger, smaller): 2. An object is located 8 cm in front of a concave mirror. If the focal length is 6 cm, locate the object and draw the ray diagram for the resulting image. C Object Type (Real or Virtual): Orientation (Upright or Inverted): Location (front or behind): Size (same, larger, smaller):
The red lines represent the incident rays, while the blue lines represent the refracted rays. The object is located between F and C, and the resulting image is real, inverted, and located beyond C.
1. The image of an object that is located 14 cm in front of a concave mirror with a focal length of 3 cm is a virtual image.Object type: Virtual Orientation: Upright Location: Behind the mirror Size: Larger Draw the ray diagram for the resulting image: 2. The image of an object that is located 8 cm in front of a concave mirror with a focal length of 6 cm is a real image.Object type: Real Orientation: Inverted Location: In front of the mirrorSize: Smaller Draw the ray diagram for the resulting image: In the above ray diagram, F is the focus, C is the center of the curvature, and P is the pole of the mirror. The red lines represent the incident rays, while the blue lines represent the refracted rays. The object is located between F and C, and the resulting image is real, inverted, and located beyond C.
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What is the pooled variance for the following two samples? sample 1: n = 8 and ss = 168; sample 2: n = 6 and ss = 120
The pooled variance is the weighted average of the variances of two or more groups, where the weights are the degrees of freedom (n-1) for each group.
To get the pooled variance for the given samples, we need to find the variance of each sample and plug in the values in the formula above. Sample 1 has n = 8
and ss = 168.
To get the variance of this sample (S1²), Plugging in the values Now let's find the variance of sample 2. It has n = 6 and ss = 120.
Therefore, the pooled variance for the given two samples is 24. The pooled variance for the given two samples is 24. The pooled variance is the weighted average of the variances of two or more groups, where the weights are the degrees of freedom (n-1) for each group. We can find the variance of each sample using the formula S² = SS/(n-1), where SS is the sum of squares and n is the sample size. Plugging in the values, we find that the variance of both samples is 24. Finally, we can use the formula Sp² = (S1²(n1-1) + S2²(n2-1))/(n1+n2-2) to find the pooled variance, which is also 24.
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A cosmic ray proton moving toward the Earth at 10. 00 × 107 m/s experiences a magnetic force of 2.10× 10−16 . What is the strength of the magnetic field if there is a 30° angle between it and the proton’s velocity?
The strength of the magnetic field is 0.7 μT.
Cosmic rays are high-energy particles that originate in space. They comprise cosmic rays of different atomic nuclei, subatomic particles such as protons, atomic nuclei like helium nuclei, and electrons, and occasionally antimatter particles such as positrons.
They also originate from galactic sources. These particles are considered primary cosmic rays because they are directly produced in cosmic ray sources.
Secondary cosmic rays, such as energetic photons, charged particles, and neutrinos, are produced when primary cosmic rays collide with atoms in the atmosphere. This creates showers of secondary particles that are observed on the Earth's surface.
Magnetic Force and Magnetic Field
A magnetic force (F) can be applied to a charged particle moving in a magnetic field (B) at a velocity v, as given by the formula:
F = qvB sin(θ)
Where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the magnetic field and the velocity of the particle.
In this problem, the magnetic force and velocity of a proton moving towards the Earth are given. The formula can be rearranged to solve for the magnetic field (B):
B = F / (qv sin(θ))
Substituting the given values:
B = 2.10 × 10^-16 N / ((1.6 × 10^-19 C)(10.00 × 10^7 m/s)sin(30°))
= 0.7 μT
Therefore, the strength of the magnetic field, if there is a 30° angle between it and the proton's velocity, is 0.7 μT.
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An important news announcement is transmitted by radio waves to people who are 300 km away and sitting next to their radios, and also by sound waves to people sitting 4.00 m from the newscaster in a newsroom. Who receives the news first? people in the newsroom both at the same time At = people next to their radios What is the difference in time At between each group of people receiving the news?
Who receives the news first and calculate the time difference between the two groups of people, we need to compare the speed of radio waves and sound waves.people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
The speed of light, which includes radio waves, is approximately 3.00 x 10^8 meters per second (m/s) in a vacuum. However, when radio waves travel through the Earth's atmosphere, they slow down slightly but the difference is negligible for this calculation.
On the other hand, the speed of sound depends on the medium through which it travels. In dry air at room temperature, the speed of sound is approximately 343 meters per second (m/s).
First, let's calculate the time it takes for the radio waves to travel a distance of 300 km:
Time taken by radio waves = Distance / Speed
= 300,000 m / (3.00 x 10^8 m/s)
≈ 1.00 x 10^(-3) seconds (or 1 millisecond)
Next, let's calculate the time it takes for sound waves to travel a distance of 4.00 meters:
Time taken by sound waves = Distance / Speed
= 4.00 m / 343 m/s
≈ 0.0117 seconds (or 11.7 milliseconds)
Therefore, people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
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A piano wire of linear mass density 0.0050 kg/m is under a tension of 1350 N. What is the wave speed in this wire? O 1040 m/s O 260 m/s O 520 m/s 130 m/s Moving to another question will save this resp
The wave speed in the piano wire, under a tension of 1350 N and linear mass density of 0.0050 kg/m, is approximately 520 m/s.
To calculate the wave speed in the piano wire, we can use the formula:
Wave speed (v) = sqrt(Tension (T) / linear mass density (μ))
Given:
Linear mass density (μ) = 0.0050 kg/m
Tension (T) = 1350 N
Substituting these values into the formula, we get:
Wave speed (v) = sqrt(1350 N / 0.0050 kg/m)
Wave speed (v) = sqrt(270,000 m²/s² / kg/m)
Wave speed (v) = sqrt(270,000) m/s
Wave speed (v) ≈ 519.62 m/s
Therefore, the wave speed in the piano wire is approximately 520 m/s.
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a stream accelerating
neutrons creates
A-electromagnetic
waves
B- an electric field
only
C-no magnetic or electric
fields
D-a magnetic field
only
When a stream of neutrons accelerates, it produces a magnetic field only. The other options are incorrect since electromagnetic waves are produced when there is a disturbance in electric and magnetic fields.
Since no electric fields are present, the option B is incorrect. In addition, there is no evidence of electromagnetic radiation which means that option A is also wrong. There is also no electrical charge to allow for the formation of an electric field. It is worth noting that an electric field is a region where an electrically charged object experiences an electric force.
As a result, option C is incorrect. Finally, a magnetic field can be produced when there is a movement of charge, like in the case of a stream of neutrons, as they are electrically neutral. When there is a movement of charge, a magnetic field is produced perpendicular to the direction of the current. As such, option D is correct. Therefore, the correct answer to the question is option D.
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(a) In brief terms, provide an account of nuclear instability, making use of the Nuclear chart "Segré chart" to illustrate your answer. (a) A particular expression of the semi-empirical formula for the binding energy of a nucleus is (in MeV): B-15.5 A-16.842) - 0.72 Z+/A!) – 19(N=Z)'/A Discuss the origin of each ten
Nuclear instability refers to the tendency of certain atomic nuclei to undergo decay or disintegration due to an imbalance between the forces that hold the nucleus together and the forces that repel its constituents.
The Segré chart, also known as the nuclear chart, is a graphical representation of all known atomic nuclei, organized by their number of protons (Z) and neutrons (N). It provides a visual representation of the stability or instability of nuclei.
The semi-empirical formula for the binding energy of a nucleus provides insights into the origin of nuclear stability. The formula is given by B = (15.5A - 16.842) - 0.72Z^2/A^(1/3) - 19(N-Z)^2/A, where B represents the binding energy of the nucleus, A is the mass number, Z is the atomic number, and N is the number of neutrons.
The terms in the formula have specific origins. The first term, 15.5A - 16.842, represents the volume term and is derived from the idea that each nucleon (proton or neutron) contributes a certain amount to the binding energy.
The second term, -0.72Z^2/A^(1/3), is the Coulomb term and accounts for the electrostatic repulsion between protons. It is inversely proportional to the cube root of the mass number, indicating that larger nuclei with more nucleons experience weaker Coulomb repulsion.
The third term, -19(N-Z)^2/A, is the symmetry term and arises from the observation that nuclei with equal numbers of protons and neutrons (N = Z) tend to be more stable. The asymmetry between protons and neutrons reduces the binding energy.
In summary, nuclear instability refers to the tendency of certain atomic nuclei to decay due to an imbalance between attractive and repulsive forces. The Segré chart provides a visual representation of nuclear stability.
The semi-empirical formula for binding energy reveals the origin of stability through its terms: the volume term, Coulomb term, and symmetry term, which account for the contributions of nucleons, electrostatic repulsion, and asymmetry, respectively.
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(a) A bullet with a mass of 2.10 g moves at a speed of 1.50 x 10³ m/s. If a tennis ball of mass 57.5 g has the same momentum as the bullet, what is its speed (in m/s)? m/s (b) Which has greater kinetic energy, the ball or the bullet? O Both have the same kinetic energy. The bullet has greater kinetic energy. O The ball has greater kinetic energy. A 7.80-g bullet moving at 540 m/s penetrates a tree trunk to a depth of 6.50 cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (Enter the magnitude.) N (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving. S A professional golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00480 s. After the collision, the ball leaves the club at a speed of 39.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club? N
(a) A bullet with a mass of 2.10 g moves at a speed of 1.50 x 10³ m/s. If a tennis ball of mass 57.5 g has the same momentum as the bullet, then its speed is 54.79 m/s.
(b) the bullet has a greater kinetic energy than the tennis ball.
(a)The average frictional force that stops the bullet is 223.6 N.
(b) Assuming the frictional force is constant, we can use Newton's second law, F = ma, to find the time it takes for the bullet to come to a stop.
Rearranging
(a) To find the speed of the tennis ball, we can use the conservation of momentum. The momentum of an object is given by the product of its mass and velocity. Since momentum is conserved in a collision, the momentum of the bullet will be equal to the momentum of the tennis ball.
Let's denote the mass of the bullet as m1 (2.10 g) and the speed of the bullet as v1 (1.50 x 10^3 m/s). The mass of the tennis ball is m2 (57.5 g), and we need to find the speed of the tennis ball, denoted as v2.The momentum of the bullet is given by p1 = m1 * v1, and the momentum of the tennis ball is given by p2 = m2 * v2. Since the momenta are equal, we can set up an equation: m1 * v1 = m2 * v2.
Plugging in the values, we have (2.10 g) * (1.50 x 10^3 m/s) = (57.5 g) * v2.
Solving for v2, we find v2 = [(2.10 g) * (1.50 x 10^3 m/s)] / (57.5 g).
Performing the calculation, v2 ≈ 54.79 m/s.
(b) The kinetic energy of an object is given by the formula KE = (1/2) * m * v^2, where m is the mass of the object and v is its velocity.Comparing the kinetic energy of the bullet and the tennis ball, we can calculate the kinetic energy for each using their respective masses and velocities.
For the bullet: KE_bullet = (1/2) * (7.80 g) * (540 m/s)^2. For the tennis ball: KE_tennis_ball = (1/2) * (55.0 g) * (39.0 m/s)^2.Performing the calculations, we find that KE_bullet ≈ 846,540 J and KE_tennis_ball ≈ 48,247 J.Thus, the bullet has a greater kinetic energy than the tennis ball.
(a) To find the average frictional force that stops the bullet, we can use the work-energy principle. The work done by the frictional force is equal to the change in kinetic energy of the bullet.
The initial kinetic energy of the bullet is given by KE_initial = (1/2) * m * v_initial^2, where m is the mass of the bullet and v_initial is its initial velocity. In this case, m = 7.80 g and v_initial = 540 m/s.
The final kinetic energy of the bullet is zero since it comes to a stop. Therefore, the work done by the frictional force is equal to the initial kinetic energy of the bullet.
The work done by the frictional force is given by W = F * d, where F is the average frictional force and d is the distance the bullet penetrates the tree trunk.
Setting W equal to KE_initial, we have F * d = KE_initial.
Rearranging the equation to solve for the average frictional force, we get F = KE_initial / d.
Plugging in the values, F = (0.5 * 7.80 g * (540 m/s)^2) / (6.50 cm).
Converting the units to N and m, F ≈ 223.6 N.
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At the starting gun, a runner accelerates at 1.9 m>s2 for 5.2 s. The runner’s acceleration is zero for the rest of the race. What is the speed of the runner (a) at t = 2.0 s, and (b) at the end of the race
At the end of the race, the time (t) is the total time of 5.2 seconds. To solve this problem, we can use the equations of motion. The equations of motion for uniformly accelerated linear motion are:
v = u + at
s = ut + (1/2)at^2
v^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
Initial velocity (u) = 0 m/s (since the runner starts from rest)
Acceleration (a) = 1.9 m/s^2
Time (t) = 5.2 s
(a) To find the speed at t = 2.0 s:
v = u + at
v = 0 + (1.9)(2.0)
v = 0 + 3.8
v = 3.8 m/s
Therefore, the speed of the runner at t = 2.0 s is 3.8 m/s.
(b) To find the speed at the end of the race:
The runner's acceleration is zero for the rest of the race. This means that the runner continues to move with a constant velocity after 5.2 seconds.
Since the acceleration is zero, we can use the equation:
v = u + at
At the end of the race, the time (t) is the total time of 5.2 seconds.
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Predict how much torque is affecting this simple motor. The area inside the rectangle is 15 cm2, the current it carries is 9 A, the magnetic field has a magnitude of 20 * 10-3 T, and the angle between the area vector and the magnetic field is 1.0 radians.
The torque affecting the simple motor can be predicted as 6 * 10⁻⁷ m² * T * sin(1.0 radians).
The torque (τ) affecting the motor can be calculated using the formula:
τ = A * B * sin(θ)
where:
A is the area of the rectangle (15 cm²),
B is the magnitude of the magnetic field (20 * 10^-3 T),
θ is the angle between the area vector and the magnetic field (1.0 radians).
Substituting the given values into the formula, we have:
τ = 15 cm² * 20 * 10^-3 T * sin(1.0 radians)
To simplify the calculation, we convert the area from cm² to m²:
τ = (15 cm² * 10^-4 m²/cm²) * 20 * 10^-3 T * sin(1.0 radians)
τ = 3 * 10^-4 m² * 20 * 10^-3 T * sin(1.0 radians)
τ = 6 * 10^-7 m² * T * sin(1.0 radians)
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An LRC circuit consists of a 19.0- μF capacitor, a resistor, and an inductor connected in series across an ac power source of variable frequency that has a voltage amplitude of 27.0 V. You observe that when the power source frequency is adjusted to 41.5 Hz, the rms current through the circuit has its maximum value of 67.0 mA. What will be the rms current irms if you change the frequency of the power source to 60.0 Hz ?
the correct option is 150.
when the frequency of the power source changes to 60.0 Hz is 0.600 A or 600 mA (approximately).
Given data,
Capacitor, C = 19.0 μF
Resistor, R = ?
Inductor, L = ?
Voltage amplitude, V = 27.0 V
Maximum value of rms current, irms = 67.0 m
A = 67.0 × 10⁻³ A
Frequency, f₁ = 41.5 Hz
Let's calculate the value of inductive reactance and capacitive reactance for f₁ using the following formulas,
XL = 2πfLXC = 1/2πfC
Substitute the given values in the above equations,
XL = 2πf₁L
⇒ L = XL / (2πf₁)XC = 1/2πf₁C
⇒ C = 1/ (2πf₁XC)
Now, substitute the given values in the above formulas and solve for the unknown values;
L = 11.10 mH and C = 68.45 μF
Now we can calculate the resistance of the LRC circuit using the following equation;
Z = √(R² + [XL - XC]²)
And we know that the impedance, Z, at resonance is equal to R.
So, at resonance, the above equation becomes;
R = √(R² + [XL - XC]²)R²
= R² + [XL - XC]²0
= [XL - XC]² - R²0
= [2πf₁L - 1/2πf₁C]² - R²
Now, we can solve for the unknown value R.
R² = (2πf₁L - 1/2πf₁C)²
R = 6.73 Ω
When frequency, f₂ = 60.0 Hz, the new value of XL = 2πf₂LAnd XC = 1/2πf₂C
We have already calculated the values of L and C, let's substitute them in the above formulas;
XL = 16.62 Ω and XC = 44.74 Ω
Now, we can calculate the impedance, Z, for the circuit when the frequency, f₂ = 60.0 Hz
Z = √(R² + [XL - XC]²)
= √(6.73² + [16.62 - 44.74]²)
= 45.00 Ω
Now, we can calculate the rms current using the following formula;
irms = V / Z = 27.0 V / 45.00 Ω = 0.600 A
Irms when the frequency of the power source changes to 60.0 Hz is 0.600 A or 600 mA (approximately).
Therefore, the correct option is 150.
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iPhones use a maximum of 2.4 A of current at 5 volts. If you charge your phone for 1.5 hours, calculate the value of charge during this time.
When charging your phone for 1.5 hours with a maximum current of 2.4 A, the value of charge transferred to the phone is 12,960 Coulombs.
Calculating the value of charge when charging your phone for 1.5 hours, we can use the formula:
Charge = Current × Time
Current (I) = 2.4 A
Time (t) = 1.5 hours
First, we need to convert the time from hours to seconds:
1.5 hours = 1.5 × 3600 seconds = 5400 seconds
Now we can calculate the charge:
Charge = 2.4 A × 5400 s = 12,960 Coulombs
Therefore, when charging your phone for 1.5 hours, the value of charge transferred to the phone is 12,960 Coulombs.
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A proton moving in a uniform magnetic field with V1 = 1.18 × 106 m/s experiences force F₁ = 1.39 × 10-16 N. A second proton with v₂ = 2.21 ×106 m/s experiences → F2: -16% N in the same field. 3.62 x 10 == What is the magnitude of B? Express your answer with the appropriate units. ► View Available Hint(s) 0 μA ? B = Value T Submit X Incorrect; Try Again Part B What is the direction of B? Give your answer as an angle measured ccw from the +x-axis. Express your answer in degrees. Previous Answers
1. The magnitude of the magnetic field is 0.38 T.
2. The direction of the magnetic field is 30 degrees counterclockwise from the +x-axis.
We can calculate the magnitude of the magnetic field using the following equation:
F = qvB sin(theta)
Where:
F is the force on the proton (1.39 × 10-16 N)
q is the charge of the proton (1.602 × 10-19 C)
v is the velocity of the proton (1.18 × 106 m/s)
B is the magnitude of the magnetic field (T)
theta is the angle between the velocity of the proton and the magnetic field (degrees)
Plugging in these values, we get:
1.39 × 10-16 N = 1.602 × 10-19 C * 1.18 × 106 m/s * B * sin(theta)
B = (1.39 × 10-16 N) / (1.602 × 10-19 C * 1.18 × 106 m/s) / sin(theta)
= 0.38 T
The direction of the magnetic field can be found using the right-hand rule. Imagine that your right hand is palm facing you, with your fingers pointing in the direction of the proton's velocity.
Your thumb will point in the direction of the magnetic field. In this case, the magnetic field is 30 degrees counterclockwise from the +x-axis.
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A full water tank in the shape of an inverted right circular cone is 14 m across the top and 7 m high. If the surface of the water in
the tank is 2 m below the top of the tank, how much work is required to pump all the water over the top of the tank? (The density
of water is 1000 kg/m, use neceleration due to gravity g = 9.8 N/kg.)
To pump all the water over the top of the tank, we need to find the volume of the water first and then use that to find the work required. The given information is as follows: Shape of the tank: Inverted right circular cone, Diameter of the top of the cone (across): 14 m, Height of the cone: 7 m, Depth of the water from the top: 2 m, Density of water: 1000 kg/m³, Acceleration due to gravity: g = 9.8 N/kg.
Formula to calculate volume of an inverted right circular cone:$$V = \frac{1}{3}πr^2h$$. Here, radius of the top of the cone, r = 14/2 = 7 m, Height of the cone, h = 7 m, Depth of the water from the top = 2 m, Height of the water, H = 7 - 2 = 5 m. So, the volume of the water in the tank is:$$V_{water} = \frac{1}{3}πr^2H$$Putting the given values,$$V_{water} = \frac{1}{3} × π × 7^2 × 5$$$$V_{water} = \frac{245}{3} π m^3$$.
To find the mass of the water, we use the formula:$$Density = \frac{mass}{volume}$$$$mass = Density × volume$$Putting the given values,$$mass = 1000 × \frac{245}{3} π$$$$mass ≈ 2.56 × 10^5 kg$$.
The work done to pump the water over the top of the tank is equal to the potential energy of the water. The formula for potential energy is:$$Potential Energy = mgh$$Here, m = mass of the water, g = acceleration due to gravity and h = height of the water above the ground. So, putting the given values,$$Potential Energy = mgh$$, $$Potential Energy = 2.56 × 10^5 × 9.8 × 5$$$$Potential Energy ≈ 1.26 × 10^7 J$$.
Therefore, the work required to pump all the water over the top of the tank is approximately equal to 1.26 × 10⁷ J.
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Consider the nuclear fusion reaction 12H+12H−>13H+11H Each fusion event releases approximately 4.03MeV of energy. How much total energy, in joules, would be released if all the deuterium atoms (12H) in a typical 0.290 kg glass of water were to undergo this fusion reaction? Assume that approximately 0.0135% of all the hydrogen atoms in the water are deuterium. energy released: Incorrect A typical human body metabolizes energy from food at a rate of about 104.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy released? time to metabolize the amount of energy released: days
To calculate the total energy released in the fusion reaction and the time it would take for a human to metabolize that energy, we need to determine the number of deuterium atoms in the given mass of water and then use the conversion factors to calculate the energy and time.
Given:
Mass of water (m) = 0.290 kg
Energy released per fusion event (E) = 4.03 MeV
Percentage of deuterium atoms in water = 0.0135%
Average human metabolic rate (P) = 104.5 W
Calculate the number of deuterium atoms in the mass of water:
Number of deuterium atoms (N) = (0.0135/100) * (6.022 × 10^23) * (0.290 kg / (2.014 g/mol))
N ≈ 1.051 × 10^19 atoms
Calculate the total energy released:
Total energy released (E_total) = N * E * (1.602 × 10^-13 J/MeV)
E_total ≈ 1.051 × 10^19 * 4.03 * (1.602 × 10^-13) J
E_total ≈ 6.78 × 10^5 J
Calculate the time to metabolize the energy:
Time (t) = E_total / P
t ≈ 6.78 × 10^5 J / 104.5 W
t ≈ 6492 s
Convert seconds to days:
t ≈ 6492 s / (24 * 60 * 60 s/day)
t ≈ 0.0752 days
The total energy released if all the deuterium atoms in a typical 0.290 kg glass of water undergo fusion is approximately 6.78 × 10^5 J.
It would take approximately 0.0752 days for a human to metabolize that amount of energy.
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A 10.9-V battery, 5.09-resistor, and a 3.5-H inductor are connected in series. After the current in the circuit has reached Is maximum valor, calculate the following (a) the power being supplied by the battery w (b) the power being delivered to the resistor w (c) the power being delivered to the Inductor w (d) the energy stored in the magnetic ned of the inductor
It can be seen that the circuit is a series circuit, hence the current passing through the circuit is same in the entire circuit. Let the current in the circuit be I. The voltage drop across the resistor is given by IR.
Hence the time derivative of current is zero, i.e., di/dt = 0.Substituting this in the above equation, we get V = I max R. This gives the value of I max = 10.9/5.09The value of I max is 2.14 A.
Power supplied by the battery; The power supplied by the battery is given by;
P = VI
Where
V = 10.9 V and
I = 2.14 A
Substituting these values, we get;
P = 23.3 W
Power delivered to the resistor; The power delivered to the resistor is given by;
P = I²R
Where
I = 2.14 A and
R = 5.09 ohm
Substituting these values, we get;
P = 24.6 W
Power delivered to the inductor; The power delivered to the inductor is given by;
P = I²L(di/dt)
I = 2.14 A,
L = 3.5 H and
di/dt = 0
Substituting these values, we get; P = 0
Energy stored in the magnetic field of the inductor; The energy stored in the magnetic field of the inductor is given by;
W = (1/2)LI²
Where
I = 2.14 A and
L = 3.5 H
Substituting these values, we get; W = 16.46 J
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As a new electrical technician, you are designing a large solenoid to produce a uniform 0.130 T magnetic field near the center of the solenoid. You have enough wire for 3000 circular turns. This solenoid must be
52.0 cm long and 2.80 cm in diameter.
What current will you need to produce the necessary field?
The magnetic field produced inside a solenoid is given asB=μ₀(n/l)I ,Where,μ₀= 4π×10^-7 T m A^-1is the permeability of free space,n is the number of turns per unit length,l is the length of the solenoid, andI is the current flowing through the wire.The solenoid has 3000 circular turns and is 52.0 cm long and 2.80 cm in diameter, and the magnetic field produced near the center of the solenoid is 0.130 T.Thus,The length of the solenoid,l= 52.0 cm = 0.52 mn= 3000 circular turns/lπd²n = 3000 circular turns/π(0.028 m)²I = ?The magnetic field equation can be rearranged to solve for current asI= (Bμ₀n/l),whereB= 0.130 Tμ₀= 4π×10^-7 T m A^-1n= 3000 circular turns/π(0.028 m)²l= 0.52 mThus,I= (0.130 T×4π×10^-7 T m A^-1×3000 circular turns/π(0.028 m)²)/0.52 m≈ 5.49 ATherefore, the current required to produce the required magnetic field is approximately 5.49 A.
The answer is a current of 386 A will be necessary. We know that the solenoid must produce a magnetic field of 0.130 T and that it has 3000 circular turns. We can determine the number of turns per unit length as follows: n = N/L, where: N is the total number of turns, L is the length
Substituting the given values gives us: n = 3000/(0.52 m) = 5769 turns/m
We can use Ampere's law to determine the current needed to produce the necessary field. According to Ampere's law, the magnetic field inside a solenoid is given by:
B = μ₀nI,where: B is the magnetic field, n is the number of turns per unit length, I is the current passing through the solenoid, μ₀ is the permeability of free space
Solving for the current: I = B/(μ₀n)
Substituting the given values gives us:I = 0.130 T/(4π×10⁻⁷ T·m/A × 5769 turns/m) = 386 A
I will need a current of 386 A to produce the necessary magnetic field.
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The dampening material in an ultrasound system is often made of
_________, and its function is to _______the pulses.
The dampening material used in an ultrasound system is often made of rubber or silicone, and its function is to absorb or reduce the intensity of the ultrasound pulses.
In an ultrasound system, the dampening material is an essential component that helps optimize the performance of the device. The material used for dampening is typically rubber or silicone, which have excellent acoustic properties. The primary purpose of the dampening material is to absorb or reduce the intensity of the ultrasound pulses emitted by the transducer.
Ultrasound pulses consist of high-frequency waves that are emitted and received by the transducer. When these pulses travel through the body, they encounter various interfaces between different tissues and organs, leading to reflections and echoes. If the ultrasound pulses were not dampened, they could bounce back and interfere with subsequent pulses, causing artifacts and reducing image quality.
By placing a layer of rubber or silicone as the dampening material in the ultrasound system, the pulses encounter resistance as they pass through the material. This resistance helps absorb or attenuate the energy of the pulses, reducing their intensity before they reach the patient's body. As a result, the echoes and reflections are less likely to interfere with subsequent pulses, allowing for clearer and more accurate imaging.
The choice of rubber or silicone as the dampening material is based on their ability to effectively absorb and attenuate ultrasound waves. These materials have properties that allow them to convert the mechanical energy of the ultrasound pulses into heat, dissipating the energy and minimizing reflection or transmission of the waves. Additionally, rubber and silicone are flexible and easily conform to the shape of the transducer, ensuring good acoustic contact and optimal dampening of the ultrasound pulses.
In conclusion, the dampening material used in an ultrasound system, typically made of rubber or silicone, serves the vital function of absorbing or reducing the intensity of ultrasound pulses. By attenuating the energy of the pulses, the dampening material helps prevent artifacts and interference, leading to improved image quality and more accurate diagnostic results.
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A thick layer of an unknown transparent liquid sits on top of water.
A ray of light in the unknown liquid encounters the surface of the water below at an incident angle of 20.0°. The ray refracts to an angle of 22.1°. If the index of refraction of water is 1.33, what is the index of refraction of the unknown liquid to three significant digits?
The index of refraction of the unknown transparent liquid is 1.21. When a ray of light goes from one medium into another, it bends or refracts at the boundary of the two media. The angle at which the incident ray approaches the boundary line is known as the angle of incidence, and the angle at which it refracts into the second medium is known as the angle of refraction.
The index of refraction for a material is a measure of how much the speed of light changes when it passes from a vacuum to the material. It may also be stated as the ratio of the speed of light in a vacuum to the speed of light in the material. It may also be used to determine the degree to which light is bent or refracted when it passes from one material to another with a different index of refraction. The following is the answer to the question:A ray of light travelling through the unknown transparent liquid has an incident angle of 20.0° and is then refracted to 22.1° upon reaching the water below.
The index of refraction for the unknown transparent liquid can be found using the following equation:
n1sinθ1 = n2sinθ2
where,θ1 is the angle of incidence,θ2 is the angle of refraction,n1 is the index of refraction of the first medium,n2 is the index of refraction of the second medium.
By substituting the values of θ1, θ2, and n1 into the above equation, we get:
n2 = n1 sin θ1 / sin θ2n1 = 1.33 (given)
n2 = n1 sin θ1 / sin θ2
= 1.33 sin 20.0° / sin 22.1°
= 1.21 to three significant figures.
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Oscillations in the elevator Gravity stretches an elastic thin wire of 1 m length by 15.5 mm as 500 g mass is attached. Determine the oscillation period, if the wire is initially stretched a little more. Which length does a pendulum thread need to have, if the pendulum should have the same period? Now put the pendulum into an elevator. The elevator accelerates and is going up: The velocity increases linearly in time during the first 3 s until reaching 24 m/s. Sketch the deflections of the pendulum versus time t in the elevator frame of reference 0.5 s before the elevator starts until 0.5 s after the start. The initial deflection is 1°. How will the deflection amplitude change qualitatively? What sort of motions of the pendulum can be observed if the elevator is going down with 9.81 m/s²?
If the elevator is going down with an acceleration of 9.81 m/s² (equal to the acceleration due to gravity), the pendulum will not experience any additional pseudo-force.
To determine the oscillation period of the elastic wire, we can use Hooke's law:
F = k * x
where F is the force, k is the spring constant, and x is the displacement.
Given that the wire is stretched by 15.5 mm (or 0.0155 m) with a 500 g (or 0.5 kg) mass attached, we can calculate the force:
F = m * g = 0.5 kg * 9.81 m/s^2 = 4.905 N
We can now solve for the spring constant:
k = F / x = 4.905 N / 0.0155 m = 316.45 N/m
The oscillation period can be calculated using the formula:
T = 2π * √(m / k)
T = 2π * √(0.5 kg / 316.45 N/m) ≈ 0.999 s
If the wire is initially stretched a little more, the oscillation period will remain the same since it depends only on the mass and the spring constant.
To find the length of the pendulum thread that would have the same period, we can use the formula for the period of a simple pendulum:
T = 2π * √(L / g)
Where L is the length of the pendulum thread and g is the acceleration due to gravity (approximately 9.81 m/s²).
Rearranging the formula, we can solve for L:
L = (T / (2π))^2 * g = (0.999 s / (2π))^2 * 9.81 m/s² ≈ 0.248 m
Therefore, the pendulum thread needs to have a length of approximately 0.248 m to have the same period as the elastic wire.
If the pendulum is put into an elevator that is accelerating upwards, the deflection of the pendulum versus time will change. Initially, before the elevator starts, the deflection will be 1°. As the elevator accelerates upwards, the deflection will increase due to the pseudo-force acting on the pendulum. The deflection will follow a sinusoidal pattern, with the amplitude gradually increasing until the elevator reaches its maximum velocity. The deflection will then start decreasing as the elevator decelerates or comes to a stop.
If the elevator is going down with an acceleration of 9.81 m/s² (equal to the acceleration due to gravity), the pendulum will not experience any additional pseudo-force. In this case, the pendulum will behave as if it is in a stationary frame of reference, and the deflection will follow a simple harmonic motion with a constant amplitude, similar to the case without any acceleration.
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A mass m = 1.81 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 86 N/m
and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion. At time t = 0 the mass is observed to be at a distance d =
0.55 m below its equilibrium height with an upward speed of vo = 4.1 m/s
The speed of the mass after a time t = 0 is 4.055 m/s.
Mass (m) = 1.81 kg
Spring Constant (k) = 86 N/m
Displacement (d) = 0.55 m
Initial Velocity (vo) = 4.1 m/s
Let's calculate the acceleration of the object using Hooke's law. According to Hooke's law,
F = -kx
where,F is the force in newtons (N)x is the displacement from the equilibrium position in meters (m)k is the spring constant in newtons per meter (N/m)
As per the problem, the displacement from the equilibrium position is d = 0.55 mForce (F) = -kx=-86 × 0.55=-47.3 N
This force acts on the mass in the upward direction. The gravitational force acting on the mass is given by
F = mg
In the given context, "m" represents the mass of the object, and "g" represents the acceleration caused by gravity. g = 9.8 m/s² (acceleration due to gravity on earth)F = 1.81 × 9.8=17.758 N
This force acts on the mass in the downward direction.
The net force acting on the mass is given by
Fnet = ma
Where a is the acceleration of the mass. We can now use Newton's second law to determine the acceleration of the mass.
a = Fnet / m = (F + (-mg)) / m= (-47.3 + (-17.758)) / 1.81= -38.525 / 1.81= -21.274 m/s² (upwards)
The negative sign shows that the acceleration is in the upward direction. Now let's find the speed of the mass after a time t.Since the mass is undergoing simple harmonic motion, we can use the equation,
x = Acos(ωt + ϕ)
Here,x is the displacement from the equilibrium position
A is the amplitude
ω is the angular frequency
t is the time
ϕ is the phase constant
At time t = 0, the mass is observed to be at a distance d = 0.55 m below its equilibrium height with an upward speed of vo = 4.1 m/s.
We can use this information to determine the phase constant. At t = 0,x = Acos(ϕ)= d = 0.55 mcos(ϕ)= d / A= 0.55 / Avo = -ωAsin(ϕ)= vo / Aωcos(ϕ)= -vo / Ax² + v₀² = A²ω²cos²(ωt) + 2Av₀sin(ωt)cos(ωt) + v₀²sin²(ωt) = A²ω²cos²(ωt) + 2Adcos(ωt) + d² - A²
Using the initial conditions, the equation becomes 0.55 = A cos ϕA(−4.1) = Aωsinϕ= −(4.1)ωcos ϕ
Squaring and adding the above two equations, we get 0.55² + (4.1ω)² = A²
Now we can substitute the known values to get the amplitude of the motion.
0.55² + (4.1ω)² = A²0.55² + (4.1 × 2π / T)² = A²
Where T is the period of the motion.
A = √(0.55² + (4.1 × 2π / T)²)
Let's assume that the object completes one oscillation in T seconds. Since we know the angular frequency ω, we can calculate the period of the motion.
T = 2π / ω = 2π / √(k / m)T = 2π / √(86 / 1.81)T = 1.281 s
Substituting the value of T, we getA = √(0.55² + (4.1 × 2π / 1.281)²)A = 1.0555 m
Now we can use the initial conditions to determine the phase constant.0.55 / 1.0555 = cos ϕϕ = cos⁻¹(0.55 / 1.0555)ϕ = 0.543 rad
Now we can use the equation for displacement,x = Acos(ωt + ϕ)= (1.0555) cos(√(k / m)t + 0.543)
Now we can differentiate the above equation to get the velocity,
v = -Aωsin(ωt + ϕ)= -(1.0555) √(k / m) sin(√(k / m)t + 0.543)When t = 0, the velocity is given byv = -(1.0555) √(k / m) sin(0.543)v = -4.055 m/s
The negative sign indicates that the velocity is in the upward direction. Thus, the speed of the mass after a time t = 0 is 4.055 m/s. Hence, the final answer is 4.055 m/s.
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In the figure(Figure 1) the coefficient of static friction between mass mA and the table is 0.43, whereas the coefficient of kinetic friction is 0.33.What value of mAmA will keep the system moving at constant speed?
To keep the system moving at a constant speed, the applied force must balance the frictional forces acting on the system.
The maximum static frictional force is given by the equation F_static = μ_static * N, where μ_static is the coefficient of static friction and N is the normal force. The kinetic frictional force is given by F_kinetic = μ_kinetic * N. Since the system is moving at a constant speed, the applied force must equal the kinetic frictional force. Therefore, to find the value of mA that keeps the system moving at a constant speed, we can set the applied force equal to the kinetic frictional force and solve for mass mA.
F_applied = F_kinetic
mA * g = μ_kinetic * (mA + mB) * g
By substituting the given values for μ_kinetic and solving for mass mA, we can find the value that keeps the system moving at a constant speed.
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A block with a mass m is floating on a liquid with a massdensity . The block has a cross-sectional area and
height . If the block is initially placed with a small vertical
displacement from the equilibrium, show that the block shows a simple harmonic motion
and then, find the frequency of the motion. Assume uniform vertical gravity with the
acceleration g
When a block with a mass of m is floating on a liquid with a mass density of ρ, the block has a cross-sectional area of A and an
acceleration
of g.
This concept can be explained in the following way:A block with a density less than that of the liquid in which it is submerged will float on the surface of the liquid with a portion of its volume submerged beneath the surface.
A floating object's volume must displace a volume of fluid equal to its own weight in order for it to remain afloat. In other words, the buoyant force on a floating object
equals the weight
of the fluid displaced by the object. The block's weight, W, must be equal to the buoyant force exerted on it, which is the product of the volume submerged, V, the liquid's density, ρ, and the gravitational acceleration, g.
As a result, we can write:W = ρVgThe volume of the
submerged block
can be expressed as hA, where h is the depth to which it is submerged. As a result, we can write V = hA. Thus, we can obtain:W = ρhAgThe block will float when its weight is less than the buoyant force exerted on it by the fluid in which it is submerged. This is when we have W < ρVg.
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. An object 1.7 cm high is held 2.5 cm from a person's cornea, and its reflected image is measured to be 0.167 cm high. Think & Prepare 1. What kind of mirror is the cornea, convex or concave?
If the image height is smaller than the object, the mirror used in the cornea is a convex mirror.
Object height (h_o) = 1.7 cm
Object distance (u) = 2.5 cm
Image height (h_i) = 0.167 cm
To find whether the mirror used is convex or concave, we need to consider the properties of the image.
When an object is placed in front of a convex mirror, the image is always with virtual and diminished. If an object is placed in front of a concave mirror, the image is always virtual or real based on the position of the mirror.
In the given scenario, the image height is smaller than the object.
Therefore we can conclude that the cornea acts as a convex mirror.
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An alien pilot of an intergalactic spaceship is traveling at 0.87c relative to a certain galaxy, in a direction parallel to its short axis. The alien pilot determines the length of the short axis of the galaxy to be 3.0 × 10^17 km. What would the length of this axis be as measured by an observer living on a planet within the galaxy?
The length of the short axis of the galaxy as measured by an observer living on a planet within the galaxy would be approximately 4.1 × 10^17 km.
The length of the short axis of the galaxy as measured by an observer living on a planet within the galaxy would be longer than the length measured by the alien pilot due to the effects of length contraction. The formula for calculating the contracted length is,
L = L0 × √(1 - v²/c²)
where:
L = contracted length
L0 = proper length (the length of the object when at rest)
v = relative speed between the observer and the object
c = speed of light
Given data:
L = 3.0 × 10¹⁷ km
v = 0.87c
Substuting the L and v values in the formula we get:
L = L0 × √(1 - v² / c²)
L0 = L / √(1 - v²/c² )
= (3.0 × 10¹⁷ km) / √(1 - (0.87c)²/c²)
= (3.0 × 10¹⁷km) /√(1 - 0.87²)
= 4.1 × 10¹⁷ km
Therefore, the length of the short axis of the galaxy as measured by an observer living on a planet within the galaxy would be approximately 4.1 × 10^17 km.
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7. Two massive objects (M1=M2=N#)kg attract each other with a force 0.128 N. What happens to the force between them if the separation between their centers is reduced to one-eighth its. original value? (Hint: F=GM2M1/R2 ) The force is now equal to : a) 3.6 N b) 42 N c) 8.2 N d) 96 N e) None of these is true
The correct answer to the question “Two massive objects (M1=M2=N#)kg attract each other with a force 0.128 N.
What happens to the force between them if the separation between their centers is reduced to one-eighth its.
original value?” is that the force is now equal to 8.2 N.
What is the gravitational force?
The force of attraction between two objects because of their masses is known as gravitational force.
The formula to calculate gravitational force is
F = Gm₁m₂/d²
where,F = force of attraction between two masses
G = gravitational constant
m₁ = mass of the first object
m₂ = mass of the second object
d = distance between the two masses.
As per the question given, the gravitational force (F) between two objects
M1=M2=N#
= N kg is 0.128 N.
Now, we are to find the new force when the distance between their centers is reduced to one-eighth of its original value.
So, we can assume that the distance is now d/8,
where d is the initial distance.
Using the formula of gravitational force and plugging the values into the formula, we have,
0.128 = G × N × N / d²
⇒ d² = G × N × N / 0.128
d = √(G × N × N / 0.128)
On reducing the distance to 1/8th, the new distance between the objects will be d/8.
Hence, we can write the new distance as d/8, which means new force F' is given as
F' = G × N × N / (d/8)²
F' = G × N × N / (d²/64)
F' = G × N × N × 64 / d²
Now, substituting the values of G, N, and d, we get
F' = 6.67 × 10^-11 × N × N × 64 / [(√(G × N × N / 0.128)]²
F' = 6.67 × 10^-11 × N × N × 64 × 0.128 / (G × N × N)
F' = 8.2 N
Thus, the new force between the two objects is 8.2 N.
Therefore, option C is correct.
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