Answer:
12 m/s²
Explanation:
Maximum acceleration is a = Aω², where A is the amplitude and ω is the frequency.
a = (0.0030 m) (10 rev/s × 2π rad/rev)²
a = 12 m/s²
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
Difference between regular and irregular object.
Hope this helps....
Good luck on your assignment.....
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.78 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.3 A. Determine the two resistances.
Answer:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
Explanation:
Ohm's Law states that:
V = IR
R = V/I
where,
R = Resistance
V = Potential Difference
I = Current
Therefore, for series connection:
Rs = Vs/Is
where,
Rs = Resistance when connected in series = R(smaller) + R(larger)
Vs = Potential Difference when connected in series = 12 V
Is = Current when connected in series = 1.78 A
Therefore,
R(smaller) + R(larger) = 12 V/1.78 A
R(smaller) + R(larger) = 6.74 Ω --------------- equation 1
R(smaller) = 6.74 Ω - R(larger) --------------- equation 2
Therefore, for series connection:
Rp = Vp/Ip
where,
Rp = Resistance when connected in parallel = [1/R(smaller) + 1/R(larger)]⁻¹
Rp = [{R(smaller) + R(larger)}/{R(smaller).R(larger)]⁻¹
Rp = R(smaller).R(larger)/[R(smaller) + R(larger)]
Vp = Potential Difference when connected in parallel = 12 V
Ip = Current when connected in parallel = 11.3 A
Therefore,
R(smaller).R(larger)/[R(smaller) + R(larger)] = 12 V/11.3 A
using equation 1 and equation 2, we get:
[6.74 Ω - R(larger)].R(larger)/6.74 Ω = 1.06 Ω
6.74 R(larger) - R(larger)² = (6.74)(1.06)
R(larger)² - 6.74 R(larger) + 7.16 = 0
solving this quadratic equation we get:
R(larger) = 5.4 Ω (OR) R(larger) = 1.3 Ω
using these values in equation 2, we get:
R(smaller) = 1.3 Ω (OR) R(smaller) = 5.4 Ω
Since, it is given in the question that R(smaller)<R(larger).
Therefore, the correct answers will be:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
A car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could. The eel has a certain amount of rotational kinetic energy when spinning at 14 spins per second. If it swam in a straight line instead, about how fast would the eel have to swim to have the same amount of kinetic energy as when it is spinning? (a) 0.5 m/s; (b) 0.7 m/s; (c) 3 m/s; (d) 5 m/s.
Answer:
(c) 3 m/s;
Explanation:
Moment of inertia of the fish eels about its long body as axis
= 1/2 m R ² where m is mass of its body and R is radius of transverse cross section of body .
= 1/2 x m x (5 x 10⁻² )²
I = 12.5 m x 10⁻⁴ kg m²
angular velocity of the eel
ω = 2 π n where n is revolution per second
=2 π n
= 2 π x 14
= 28π
Rotational kinetic energy
= 1/2 I ω²
= .5 x 12.5 m x 10⁻⁴ x(28π)²
= 4.8312m J
To match this kinetic energy let eel requires to have linear velocity of V
1 / 2 m V² = 4.8312m
V = 3.10
or 3 m /s .
In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.220 m from the center. He then mounts the tire in a vertical plane and notes that the radius vector to the spot is at an angle of 30.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.90 rad/s2. (Assume the spot's position is initially positive, and assume the angular acceleration is in the positive direction).A) What is the angular speed of the wheel after 1.30 s?
B) What is the tangential speed of the spot after 1.30 s?
C) What is the magnitude of the total acceleration of the spot after 1.30 s?
D) What is the angular position of the spot after 1.30 s?
Answer:
a) The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex], b) The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex], c) The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex], d) The angular position of the spot is 2.130 radians (122.011°).
Explanation:
a) Given that tire accelerates at constant rate, final angular speed can be predicted by using the following formula:
[tex]\omega = \omega_{o} + \alpha \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\Delta t[/tex] - Time, measured in seconds.
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex] (starts at rest), [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex] and [tex]\Delta t = 1.30\,s[/tex], the final angular speed is:
[tex]\omega = 0\,\frac{rad}{s} + \left(1.90\,\frac{rad}{s^{2}} \right) \cdot (1.30\,s)[/tex]
[tex]\omega = 2.47\,\frac{rad}{s}[/tex]
The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex].
b) The tangential speed of the spot is the product of the distance between the center of the wheel and spot. That is:
[tex]v = r \cdot \omega[/tex]
Where r is the distance between the center of the wheel and spot. The tangential speed of the spot after 1.30 seconds is:
[tex]v = (0.220\,m)\cdot \left(2.47\,\frac{rad}{s} \right)[/tex]
[tex]v = 0.543\,\frac{m}{s}[/tex]
The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex].
c) The magnitude of the total acceleration of the spot is the magnitude of the vectorial sum of radial and tangential accelerations (both components are perpendicular to each other), which is determined by the Pythagorean theorem, that is:
[tex]a = \sqrt{a_{r}^{2} + a_{t}^{2}}[/tex]
Where [tex]a_{r}[/tex] and [tex]a_{t}[/tex] are the radial and tangential accelerations.
[tex]a = r\cdot \sqrt{\omega^{4} + \alpha^{2}}[/tex]
If [tex]r = 0.220\,m[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], then, the resultant acceleration is:
[tex]a = (0.220\,m)\cdot \sqrt{\left(2.47\,\frac{rad}{s} \right)^{4}+\left(1.90\,\frac{rad}{s^{2}} \right)^{2}}[/tex]
[tex]a \approx 1.406\,\frac{m}{s^{2}}[/tex]
The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex].
d) Let be 30° (0.524 radians) the initial angular position of the spot with respect to center. The final angular position is determined by the following equation of motion:
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta - \theta_{o})[/tex]
Final angular position is therefore cleared:
[tex]\theta - \theta_{o} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
[tex]\theta = \theta_{o} + \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
Given that [tex]\theta_{o} = 0.524\,rad[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], the angular position of the spot after 1.30 seconds is:
[tex]\theta = 0.524\,rad +\frac{\left(2.47\,\frac{rad}{s} \right)^{2} - \left(0\,\frac{rad}{s}\right)^{2}}{2\cdot \left(1.90\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]\theta = 2.130\,rad[/tex]
[tex]\theta = 122.011^{\circ}[/tex]
The angular position of the spot is 2.130 radians (122.011°).
In the rotational motion of an object, the angular acceleration is always towards the center, and the further discussion is as follows:
Rotational motion:The tangential acceleration of the object keeps changing its direction as the object rotates, always directed toward the tangent of the circle passing through the position of the object.
Radius of the spot, r = 0.220
minitial angle from the horizontal, θ = 30°
angular acceleration, α = 1.9 rad/s²
(a) from the first equation of motion we get:
ω = ω₀ + αt where
ω is the final angular speed
ω₀ is the initial angular speedand
t is the time = 1.3sω = 1.9×1.3 rad/sω = 2.47 rad/s
(b) tangential speed (v) is given by:
v = r×ωv = 0.220×2.47 m/sv = 0.5434 m/s
(c) The instantaneous tangential acceleration is given by:
[tex]a_t[/tex] = rω²so the resultant acceleration will be:
[tex]a=\sqrt{a_t^2+\alpha^2}\\\\a =\sqrt{r^2\omega^4+\alpha^2}\\\\a= \sqrt{(0.220)^2(2.47)^4+(1.9)^2}\\\\a = 1.4 \ \frac{m}{s^2}[/tex]
(d)
The angular displacement is given by:
θ = θ₀t + ¹/₂αt²θ₀ = 30° = 0.524
rad θ = 0.524×1.3 + ¹/₂×1.9×1.3²θ = 2.286 radθ = 131°
Following are the solution for points:
For a)
The angular speed is 2.47 rad/s
For b)
The tangential speed is 0.5434 m/s
For c)
Total acceleration is 1.4 m/s²
For d)
The final angular position is 131°
Learn more about rotational here:
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A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?
Answer:
About [tex]1.795 \times 10^{-3}[/tex] seconds
Explanation:
[tex]\Delta p=F \Delta t[/tex], where delta p represents the change in momentum, F represents the average force, and t represents the change in time.
The change of velocity is:
[tex]37-(-27.1)=64.1m/s[/tex]
Meanwhile, the mass stays the same, meaning that the change in momentum is:
[tex]64.1\cdot 0.14kg=8.974[/tex]
Plugging this into the equation for impulse, you get:
[tex]8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s[/tex]
Hope this helps!
Identify the following as combination, decomposition, replacement, or ion exchange reactions: Al(s) + 3 Cl2(g) → 2 AlCl3(s) Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l
Answer:
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction.
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction.
Explanation:
A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:
A + B → AB
where A and B represent any two chemical substances.
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction because a single compound forms from two or more reacting species.
Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.
The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.
In general, this type of reaction can be expressed as:
AB + CD ⇒ AD + CD
In the reaction:
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 9.50 m2. The total power it absorbs from the star is 4400 W. Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K).
Answer:
T = 300.6K
Explanation:
In this problem, since the asteroid is an ideal absorber, we can approximate it to a black body, and use Stefan's law
P = σ A e T⁴
where P is the absorbed power, A the area of the asteroid, and the emissivity that for a black body is worth 1 and sigma the Stefan_boltzmann constant 5,670 10⁻⁸ W / m² K⁴
they ask us for the temperature of the asteroid
T = [tex]\sqrt[4]{(P / \sigma A e)}[/tex]
let's calculate
T = (4400 / (5,670 10⁻⁸ 9.50 1)
T =(81.6857 108)
T = 3,006 102 K
T = 300.6K
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 2.47×1014 s−1. g
Answer:
E = 0.965eV
Explanation:
In order to calculate the minimum energy needed to eject the electrons you use the following formula:
[tex]E=h \nu[/tex] (1)
h: Planck' constant = 6.626*10^{-34}J.s
v: threshold frequency = 2.47*10^14 s^-1
You replace the values of v and h in the equation (1):
[tex]E=(6.262*10^{-34}J.s)(2.47*10^{14}s^{-1})=1.54*10^{-19}J[/tex]
In electron volts you obtain:
[tex]1.54*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=0.965eV[/tex]
The minimum energy needed is 0.965eV
A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
For more information please refer to the link below
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Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?
Answer:
ΔV = -1321.73V
Explanation:
The change in potential along the path from C to D is given by the following expression:
[tex]\Delta V=-\int_a^bE dr[/tex] (1)
E: electric field produced by a charge at a distance of r
a: distance to the sphere at position C = 0.021m
b: distance to the sphere at position D = 0.055m
The electric field is given by:
[tex]E=k\frac{Q}{r^2}[/tex] (2)
Q: charge of the sphere = 5nC = 5*10^-9C
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
You replace the expression (2) into the equation (1) and solve the integral:
[tex]\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b[/tex] (3)
You replace the values of a and b:
[tex]\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V[/tex]
The change in the potential along the path C-D is -1321.73V
What is the on ohooke benden
er ord power
What is the main difference between work, power and energy
Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
What will happen to an astronaut when the jets produce these four forces
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 60 W incandescent bulb can be replaced by a 12 W LED bulb. Both produce 800 lumens of light. Assuming the cost of electricity is $0.29 per kilowatt-hour, how much does it cost (in dollars) to run the LED bulb for one year if it runs for four hours a day?
Answer:
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
Explanation:
Given;
Power of Led bulb P = 12 W
Rate r = $0.29 per kilowatt-hour
Time = 4 hours per day
The number of hours used in a year is;
time t = 4 hours per day × 365 days per year
t = 1460 hours
The energy consumption of Led bulb in a year is;
E = Pt
E = 12 W × 1460 hours
E = 17520 watts hour
E = 17.52 kilowatt-hour
The cost of the energy consumption is;
C = E × rate = Er
C = 17.52 × $0.29
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the maximum rise in temperature of the water if the bird gives it all of its mechanical energy
Answer:
0.00185 °C
Explanation:
From the question,
The potential energy of the bird = heat gained by the water in the fish tank.
mgh = cm'(Δt)................... Equation 1
Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.
make Δt the subject of the equation
Δt = mgh/cm'............... Equation 2
Given: m = 1 kg, h = 40 m, m' = 50.5 kg
constant: g = 9.8 m/s², c = 4200 J/kg.K
Substitute into equation 2
Δt = 1(40)(9.8)/(50.5×4200)
Δt = 392/212100
Δt = 0.00185 °C
C2B.7Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil. (a) What is the earth's speed just before the anvil hits
Complete Question
C2B.7
Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil.
(a) What is the earth's speed just before the anvil hits?
b) How long would it take the earth to travel [tex]1.0 \mu m[/tex] (about a bacterium's width) at this speed?
Answer:
a
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
b
[tex]t = 9.95 *10^{15} \approx 10 *10^{15} \ s[/tex]
Explanation:
From the question we are told that
The mass of the anvil is [tex]m_a = 60\ kg[/tex]
The speed at which it hits the ground is [tex]v = 10 \ m/s[/tex]
Generally the mass of the earth has a value [tex]m_e = 5972*10^{24} \ kg[/tex]
Now according to the principle of momentum conservation
[tex]P_i = P_f[/tex]
Where [tex]P_i[/tex] is the initial momentum which is zero given that both the anvil and the earth are at rest
Now [tex]P_f[/tex] is the final momentum which is mathematically represented as
[tex]P_f = m_a * v + m_e * v_1[/tex]
So
[tex]0 = m_a * v + m_e * v_1[/tex]
substituting values
[tex]0 = 60 * 10 + 5.972 *10^{24} * v_1[/tex]
=> [tex]v_1 = -1.0*10^{-22} \ m/s[/tex]
Here the negative sign show that it is moving in the opposite direction to the anvil
The magnitude of the earths speed is
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
The time it would take the earth is mathematically represented as
[tex]t = \frac{d}{|v_1|}[/tex]
substituting values
[tex]t = \frac{1.0*10^{-6}}{1.0 *10^{-22}}[/tex]
[tex]t = 10 *10^{15} \ s[/tex]
You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
[tex]F_{e} = F_{m}[/tex]
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
A bicycle rider has a speed of 20.0 m/s at a height of 60 m above sea level when he begins coasting down hill. Sea level is the zero level for measuring gravitational potential energy. Ignoring friction and air resistance, what is the rider's speed when he coasts to a height of 18 m above sea level?
Answer:
The rider's speed will be approximately 35 m/s
Explanation:
Initially the rider has kinetic and potential energy, and after going down the hill, some of the potencial energy turns into kinetic energy. So using the conservation of energy, we have that:
[tex]kinetic_1 + potencial_1 = kinetic_2 + potencial_2[/tex]
The kinetic and potencial energy are given by:
[tex]kinetic = mass * speed^2 / 2[/tex]
[tex]potencial = mass * gravity * height[/tex]
So we have that:
[tex]m*v^2/2 + mgh = m*v'^2/2 + mgh'[/tex]
[tex]20^2/2 + 9.81*60 = v'^2/2 + 9.81*18[/tex]
[tex]v'^2/2 + 176.58 = 788.6[/tex]
[tex]v'^2/2 = 612.02[/tex]
[tex]v'^2 = 1224.04[/tex]
[tex]v' = 34.99\ m/s[/tex]
So the rider's speed will be approximately 35 m/s
New evidence increasingly emphasizes that __________.
Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of expansion twice as large as the second bar. How does the original volume of the first bar compare to the original volume of the second bar
Answer:
The original volume of the first bar is half of the original volume of the second bar.
Explanation:
The coefficient of cubic expansivity of substances is given by;
γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
Given: two metal bars with equal change in volume, equal change in temperature.
Let the volume of the first metal bar be represented by [tex]V_{1}[/tex], and that of the second by [tex]V_{2}[/tex].
Since they have equal change in volume,
Δ[tex]V_{1}[/tex] = Δ[tex]V_{2}[/tex] = ΔV
For the first metal bar,
2γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
⇒ Δθ = ΔV ÷ (2γ[tex]V_{1}[/tex])
For the second metal bar,
γ = ΔV ÷ ([tex]V_{2}[/tex]Δθ)
⇒ Δθ = ΔV ÷ ([tex]V_{2}[/tex]γ)
Since they have equal change in temperature,
Δθ of first bar = Δθ of the second bar
ΔV ÷ (2γ[tex]V_{1}[/tex]) = ΔV ÷ ([tex]V_{2}[/tex]γ)
So that;
(1 ÷ 2[tex]V_{1}[/tex]) = (1 ÷ [tex]V_{2}[/tex])
2[tex]V_{1}[/tex] = [tex]V_{2}[/tex]
[tex]V_{1}[/tex] = [tex]\frac{V_{2} }{2}[/tex]
Thus, original volume of the first bar is half of the original volume of the second bar.
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.24 m from the slits. The first bright fringe is located 3.30 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 509 nm
Explanation:
In order to calculate the wavelength of the light you use the following formula:
[tex]y=m\frac{\lambda D}{d}[/tex] (1)
where
y: distance of the mth fringe to the central peak = 3.30 mm = 3.30*10^-3 m
m: order of the bright fringe = 1
D: distance from the slits to the screen = 3.24 m
d: distance between slits = 0.500 mm = 0.500*10^-3 m
You first solve the equation (1) for λ, and then you replace the values of the other parameters:
[tex]\lambda=\frac{dy}{mD}\\\\\lambda=\frac{(0.500*10^{-3}m)(3.30*10^{-3}m)}{(1)(3.24m)}=5.09*10^7m\\\\\lambda=509*10^{-9}m=509nm[/tex]
The wavelength of the light is 509 nm
Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ball from the same height. Use the Energy Interaction Model to predict which ball will hit the ground with greater speed.
Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]
[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]
Here, initial velocity and final potential energy is zero.
[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]
Put the value into the formula
[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]
[tex]v_{f}^2=2\times9.8\times1.5[/tex]
[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]
[tex]v_{f}=5.42\ m/s[/tex]
Hence, the greater speed of the ball is 5.42 m/s.
Sea Food is a rich source of ______. *
Answer:
Sea Food is a rich source of ______. *
Explanation:
Seafood is a rich source of minerals, such as iron, zinc, iodine, magnesium, and potassium.
Answer:
Sea food is a rich source of protein
Explanation:
You should eat fish (if you are not vegan/ vegetarian ofc) at least 2 times a week and 1 has to be oily
A wire with mass 90.0g is stretched so that its ends are tied down at points 88.0cm apart. The wire vibrates in its fundamental mode with frequency 80.0Hz and with an amplitude of 0.600cm at the antinodes.a) What is the speed of propagation of transverse waves in the wire?b) Compute the tension in the wire.
Answer:
a) V = 140.8 m/s
b) T = 2027.52 N = 2.03 KN
Explanation:
a)
The formula for the speed of the wave is given as follows:
f₁ = V/2L
V = 2f₁L
where,
V = Speed of Wave = ?
f₁ = Fundamental Frequency = 80 Hz
L = Length of Wire = 88 cm = 0.88 m
Therefore,
V = (2)(80 Hz)(0.88 m)
V = 140.8 m/s
b)
Another formula for the speed of wave is:
V = √T/μ
V² = T/μ
T = V²μ
where,
T = Tension in String = ?
μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m
μ = 0.1 kg/m
Therefore,
T = (140.8 m/s)²(0.1 kg/m)
T = 2027.52 N = 2.03 KN