Answer:
1.2 s
Explanation:
Given:
v₀ = 8.0 m/s
v = -4.0 m/s
a = -10 m/s²
Find: t
v = at + v₀
(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)
t = 1.2 s
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?
Answer:
The apparent weight of the pilot is 3311 N
Explanation:
Given;
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]
Real weight of the pilot;
[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]
Apparent weight - Real weight of pilot = weight due to centripetal force
[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]
Therefore, the apparent weight of the pilot is 3311 N
An ice skater is in a fast spin with her arms held tightly to her body. When she extends her arms, which of the following statements in NOT true?
A. Het total angular momentum has decreased
B. She increases her moment of inertia
C. She decreases her angular speed
D. Her moment of inertia changes
Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the mass of the rock? (Show work).
Answer:
1 kgExplanation:
Check the diagram attached below for the diagram.
Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark
Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.
Moment = Force * perpendicular distance
The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction
Clockwise moment = 1kg * 25 = 25kg/cm
Anticlockwise moment = W * 25cm = 25W kg/cm
Equating both moments of forces
25W = 25
W = 25/23
W = 1 kg
The mass of the rock is also 1 kg
A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?
Answer:
95.5 m
Explanation:
The displacement is the position of the ending point relative to the starting point.
In this case, the magnitude of the displacement is the diameter of the circular track.
d = 300 m / π
d ≈ 95.5 m
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
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The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through the center of gravity and weigh the two parts. What will you discover?
Answer:
I believe it is they will weigh the same
Explanation:
Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics
The head side is heavier than the handle side. - this will be discovered.
What is center of gravity of a object?Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.
We know that center of gravity is close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.
So, we can conclude that the head side of the ax is heavier than the handle side of it.
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A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2 cm^2. A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 9.20 N⋅m^2/C, what area of the sheet is enclosed by the Gaussian surface?
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.
Charge and Charge DensityA certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.
Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.
[tex]\sigma = \dfrac {Q}{A}[/tex]
[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]
[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]
Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].
Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.
[tex]Q' = \sigma A'[/tex]
Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.
For the closed Gaussian Surface, Flux is given below.
[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]
Hence the charge can be written as,
[tex]Q' = \phi\epsilon_\circ[/tex]
So the charge can be given as below.
[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]
Then the surface area of the Gaussian surface is given below.
[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]
Substituting the values in the above equation,
[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]
[tex]A' =0.006076\;\rm m^2[/tex]
[tex]A' = 60.76 \;\rm cm^2[/tex]
Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.
To know more about the charge and charge density, follow the link given below.
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The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an overall angular magnification of 300
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is [tex]f_e = 0.027 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]
This negative sign shows the the microscope is diverging light
The angular magnification is [tex]m = 300[/tex]
The distance between the objective and the eyepieces lenses is [tex]Z = 19 \ cm = 0.19 \ m[/tex]
Generally the magnification is mathematically represented as
[tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]
Where [tex]f_e[/tex] is the eyepiece focal length of the microscope
Now making [tex]f_e[/tex] the subject of the formula
[tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]
substituting values
[tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]
[tex]f_e = 0.027 \ m[/tex]
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
time =
Answer:
The waiting time is [tex]t_w = 3 .47 \ s[/tex]
Explanation:
From the question we are told that
The height of the hot air balloon above the ground is [tex]d = 50 \ m[/tex]
The distance of the balloon from the target is [tex]l = 100 \ m[/tex]
The velocity of the balloon is [tex]v = 15 \ m/s[/tex]
Generally the time it will take to reach the ground is
[tex]t = \sqrt{2 * \frac{d}{g} }[/tex]
substituting values
[tex]t = \sqrt{2 * \frac{50}{9.8} }[/tex]
[tex]t = 3.2 \ s[/tex]
The distance that is covered at time with the given velocity is mathematically evaluated as
[tex]z = v * t[/tex]
substituting values
[tex]z = 15 * 3.2[/tex]
[tex]z = 48 \ m[/tex]
This implies that for the balloon moving at a velocity (v) to hit the target it must be dropped at this distance (z)
Now the distance the balloonist has to wait before dropping in order to hit the target is
[tex]A = d - z[/tex]
substituting values
[tex]A = 100 - 48[/tex]
[tex]A = 52 \ m[/tex]
This implies that the time the balloonist has to wait is
[tex]t_w = \frac{A}{v}[/tex]
substituting values
[tex]t_w = \frac{52}{15}[/tex]
[tex]t_w = 3 .47 \ s[/tex]
The potential energy function
U(x,y)=A[(1/x2) + (1/y2)] describes a conservative force, where A>0.
Derive an expression for the force in terms of unit vectors i and j.
Answer:
[tex]F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex]
Explanation:
You have the following potential energy function:
[tex]U(x,y)=A[\frac{1}{x^2}+\frac{1}{y^2}}][/tex] (1)
A > 0 constant
In order to find the force in terms of the unit vectors, you use the gradient of the potential function:
[tex]\vec{F}=\bigtriangledown U(x,y)=\frac{\partial}{\partial x}U\hat{i}+\frac{\partial}{\partial y}U\hat{j}[/tex] (2)
Then, you replace the expression (1) into the expression (2) and calculate the partial derivatives:
[tex]\vec{F}=A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]} \hat{i}+A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]\hat{j}\\\\\vec{F}=A(-2x^{-3})\hat{i}+A(-2y^{-3})\hat{j}\\\\F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex](3)
The result obtained in (3) is the force expressed in terms of the unit vectors, for the potential energy function U(x,y).
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s
Answer:
389 kg
Explanation:
The computation of mass is shown below:-
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where K indicates spring constant
m indicates mass
For the new time period
[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]
Now, we will take 2 ratios of the time period
[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]
[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]m' = \frac{0.500}{0.5625}[/tex]
= 0.889 kg
Since mass to be sum that is
= 0.889 - 0.500
0.389 kg
or
= 389 kg
Therefore for computing the mass we simply applied the above formula.
The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.
Given :
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.
The formula of the time period is given by:
[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex] ---- (1)
where m is the mass and K is the spring constant.
The new time period is given by:
[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex] ---- (2)
where m' is the total mass after the addition and K is the spring constant.
Now, divide equation (1) by equation (2).
[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]
Now, substitute the known terms in the above expression.
[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]
Simplify the above expression in order to determine the value of m'.
[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]
m' = 0.889 Kg
Now, the mass added to the object to change the period to 2.00 s is given by:
m" = 0.889 - 0.500
m" = 0.389 Kg
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A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. Applying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm
Answer:
6 atm
Explanation:
PV = PV
(4 atm) (3 L) = P (2 L)
P = 6 atm
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacitor?Now a conductor is inserted into the capacitor. The thickness of the conductor is 1/3 the distance between the plates of the capacitor and is centered inbetween the plates of the capacitor.b) What is the charge on the plates of the capacitor?c) What is the capacitance of the capacitor with the conductor in place?d) What is the energy stored in the capacitor with the conductor in place?
Answer:
a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]
b) Q = 45 µC
c) C' = 1.5 μF
d) [tex]E = 6.75 *10^{-4} J[/tex]
Explanation:
Capacitance, C = 1 µF
Charge on the plates, Q = 45 µC
a) Energy stored in the capacitor is given by the formula:
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]
b) The charge on the plates of the capacitor will not change
It will still remains, Q = 45 µC
c) Electric field is non zero over (1-1/3) = 2/3 of d
From the relation V = Ed,
The voltage has changed by a factor of 2/3
Since the capacitance is given as C = Q/V
The new capacitance with the conductor in place, C' = (3/2) C
C' = (3/2) * 1μF
C' = 1.5 μF
d) Energy stored in the capacitor with the conductor in place
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation:
When a potential difference of 12 V is applied to a wire 7.2 m long and 0.35 mm in diameter the result is an electric current of 2.1 A. What is the resistivity of the wire?
Answer:
7.63 x 10^-8ohmm
Explanation:
resistivity of the wire = 7.63 x 10^-8ohmm
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
1. As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty
Answer:
It's more likely that the trailer is heavily loaded
Explanation:
Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 N.m. When the dipole reaches equilibrium its potential energy is equal to -0.2 J. What was the initial angle between the direction of the dipole moment and the direction of the electric field?
Answer:
θ = 180
Explanation:
When an electric dipole is placed in an electric field, there is a torque due to the electric force
τ = p x E
by rotating the dipole there is a change in potential energy
ΔU = ∫ τ dθ
ΔU = p E (cos θ₂ - cos θ₁)
when the dipole starts from an angle to the equilibrium position for θ = 0
ΔU = pE (cos θ - cos 0)
cos θ = 1 + DU / pE)
let's apply this expression to our case, the change in potential energy is ΔU = -0.2J
let's calculate
cos θ = 1 -0.2 / 0.1
cos θ = -1
θ = 180
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is 25.1 m/s, how much should Raquel bank the turn so that a vehicle traveling at the posted speed limit can make the turn without relying on the frictional force between the tires and the road
Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
Two space ships collide in deep space. Spaceship P, the projectile, has a mass of 4M,
while the target spaceship T has a mass of M. Spaceship T is initially at rest and the
collision is elastic. If the final velocity of Tis 8.1 m/s, what was the initial velocity of
P?
Answer:
The initial velocity of spaceship P was u₁ = 5.06 m/s
Explanation:
In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:
[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]
Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:
V₂ = Final Speed of Spaceship T = 8.1 m/s
m₁ = mass of spaceship P = 4 M
m₂ = mass of spaceship T = M
u₁ = Initial Speed of Spaceship P = ?
u₂ = Initial Speed of Spaceship T = 0 m/s
Using these values in the given equation, we get:
[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]
8.1 m/s = (8 M/5 M)u₁
u₁ = (5/8)(8.1 m/s)
u₁ = 5.06 m/s
Objects A and B are both positively charged. Both have a mass of 900 g, but A has twice the charge of B. When A and B are placed 30.0 cm apart, B experiences an electric force of 0.870 N.
How large is the force on A?
What is the charge on qA and qB?
If the objects are released, what is the initial acceleration of A?
Answer:
- Force on A = 0.870N
- charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- a = 0.966 m/s^2
Explanation:
- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.
Then, the force on A is 0.870N
- In order to calculate the charge of both objects, you use the following formula:
[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
r: distance between the objects = 30.0cm = 0.30m
A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.
You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.
[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]
Then, you have:
charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:
[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]
m: mass of the object A = 900g = 0.900kg
[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]
The acceleration of A is 0.966m/s^2
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.