Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
A small car and an SUV are at a stoplight. The car has a mass equal to half that of the SUV, and the SUV's engine can produce a maximum force equal to twice that of the car. When the light turns green, both drivers floor it at the same time. Which vehicle pulls ahead of the other vehicle after a few seconds?
Complete Question
A small car and an SUV are at a stoplight. The car has amass equal to half that of the SUV, and the SUV's engine can produce a maximum force equal to twice that of the car. When the light turns green, both drivers floor it at the same time. Which vehicle pulls ahead of the other vehicle after a few seconds?
a) It is a tie.
b) The SUV
c) The car
Answer:
The correct option is a
Explanation:
From the question we are told that
The mass of the car is [tex]m_c[/tex]
The force of the car is F
The mass of the SUV is [tex]m_s = 2 m_c[/tex]
The force of the SUV is [tex]F_s = 2 F[/tex]
Generally force of the car is mathematically represented as
[tex]F= m_ca_c[/tex]
[tex]a_c[/tex] is acceleration of the car
Generally force of the car is mathematically represented as
[tex]F_s = m_s * a_s[/tex]
[tex]a_s[/tex] is acceleration of the SUV
=> [tex]2 F = 2 m_c a_s[/tex]
[tex]F = m_c a_s[/tex]
=> [tex]m_c a_s = m_ca_c[/tex]
So [tex]a_s = a_c[/tex]
This means that the acceleration of both the car and the SUV are the same
The air flowing over the top of the wing travels
in the same amount of time than the air
flowing beneath the wing.
Answer: Short Answer: NO ( In Most Cases)
Explanation:
If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.
pls what is the difference between Ac power and dc power
Answer:
The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."
I really need help with this question someone plz help !
Answer:weight
Explanation:weight
Which factor caused higher oil prices to directly lead to inflation?
It increased demand for cars, leading to higher automobile prices.
Companies passed on production and transportation costs to consumers.
The government began to print more money.
Gas prices declined too quickly, leading to oversupply
Answer: B, Companies passed on production and transportation costs to consumers
Explanation:
A higher oil price occurred when companies passed on production and transportation costs to consumers.
Cause of high price of oilThe oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.
The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.
Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.
Learn more about inflations here: https://brainly.com/question/1082634
Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric force. A 3C charge acted on by a 5N electric force. A 4C charge acted on by a 6N electric force. A 2C charge acted on by a 6N electric force. A 3C charge acted on by a 3N electric force. A 4C charge acted on by a 2N electric force. All of the above are experiencing electric fields with the same magnitude
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) A 2C charge acted on by a 4 N electric force
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) A 3 C charge acted on by a 5 N electric force
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) A 4 C charge acted on by a 6 N electric force
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) A 2 C charge acted on by a 6 N electric force
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) A 3 C charge acted on by a 3 N electric force
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) A 4 C charge acted on by a 2 N electric force
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
A 2 C charge acted on by a 6 N electric force
1. Which of the following is NOT a vector quantity? (a) Displacement. (b) Energy. (c) Force. (d) Momentum. (e) Velocity.
Answer:
B. energy
Explanation:
A vector has direction.
Energy does not have a direction.
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
Answer:
[tex]\Delta p=1.3475\ kg-m/s[/tex]
Explanation:
The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-
We represent
The ball mass = m = 275 g = 0.275 kg
Thus it goes to the floor and resurfaces upward.
The ball hits the ground at 3.30 m/s speed that is
u = -3.30 m/s which represents the Negative since the ball hits the ground)
It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)
[tex]\Delta p=p_f-p_i[/tex]
[tex]\Delta p=m(v-u)[/tex]
[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]
[tex]\Delta p=1.3475\ kg-m/s[/tex]
small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? It is impossible to tell since the velocities are not given. The truck experiences the greater average force. It is impossible to tell since the masses are not given. The small car and the truck experience the same average force. The small car experiences the greater average force.
Answer:
The correct option is D: "The small car and the truck experience the same average force."
Explanation:
The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.
a vector has components x=6 m and y=8 m. what is its magnitude and direction?
Answer: 10m
Explanation:
The magnitude of the vector would be 10
[tex]\sqrt{6^{2}+8^{2} } =10[/tex]
A river flows due south with a speed of 5.00 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4.00 m/s due east. The river is 780 m wide. Part A What is the magnitude of his velocity relative to the earth
Answer:
6.4 m/s
Explanation:
From the question, we are given that
Speed of the river, v(r) = 5 m/s
velocity relative to the water, v(w) = 4 m/s
Width of the river, d = 780 m
The magnitude of his velocity relative to the earth is v(m)
v(m) can be gotten by using the relation
[v(m)]² = [v(w)]² + [v(r)]²
[v(m)]² = 4² + 5²
[v(m)]² = 16 + 25
[v(m)]² = 41
v(m) = √41
v(m) = 6.4 m/s
thus, the magnitude of the velocity relative to earth is 6.4 m/s
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 80 mph. How many minutes does he save?
Answer:
t = 25.5 min
Explanation:
To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.
[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]
Next, you calculate the difference between both times t1 and t2:
[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]
This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:
[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]
hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph
What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT magnetic field if the maximum value of the induced emf is to equal 8.0 V
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.
Calculation of the minimum frequency:Since
number of turns, N = 200 turns
cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
the magnitude of magnetic field strength, B = 30 x 10⁻³ T
the maximum value of the induced emf, E = 8 V
Now
Maximum induced emf should be
E = NBAω
here,
ω is angular velocity (ω = 2πf)
Now
E = NBA2πf
here,
f is the minimum frequency
So,
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz.
Learn more about frequency here: https://brainly.com/question/24470698
A kicked ball rolls across the grass and eventually comes to a stop in 4.0 sec. When the ball was kicked, its initial velocity was 20 mi/ hr. What is the acceleration of the ball as it rolls across the grass?
Answer:
-2.24 m/s²
Explanation:
Given:
v₀ = 20 mi/hr = 8.94 m/s
v = 0 m/s
t = 4.0 s
Find: a
v = v₀ + at
0 m/s = 8.94 m/s + a (4.0 s)
a = -2.24 m/s²
You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power is [tex]P_p = 4.6 \ W[/tex]
Explanation:
From the question we are told that
The power delivered is [tex]P_{s} = 1.15 \ W[/tex]
Let it resistance be denoted as R
The resistors are connected in series so the equivalent resistance is
[tex]R_{eqv} = R+ R = 2 R[/tex]
Considering when it is connected in series
Generally power is mathematically represented as
[tex]P_s = V * I[/tex]
Here I is the current which is mathematically represented as
[tex]I = \frac{V}{2R}[/tex]
The power becomes
[tex]P_s = V * \frac{V}{2R}[/tex]
[tex]P_s = \frac{V^2}{2R}[/tex]
substituting value
[tex]1.15 = \frac{V^2}{2R}[/tex]
Considering when resistance is connected in parallel
The equivalent resistance becomes
[tex]R_{eqv} = \frac{R}{2}[/tex]
So The current becomes
[tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]
And the power becomes
[tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]
substituting values
[tex]P_p = 4 * 1.15[/tex]
[tex]P_p = 4.6 \ W[/tex]
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine their mass by measuring the period of oscillation when sitting in a chair connected to a spring. Suppose a chair is connected to a spring with a spring constant of 600 N/m. If the empty chair oscillates with a period of 0.9s, what is the mass of an astronaut who oscillates with a period of 2.0 s while sitting in the chair
Answer:
ma = 48.48kg
Explanation:
To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:
[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex] (1)
mc: mass of the chair
k: spring constant = 600N/m
T: period of oscillation of the chair = 0.9s
You solve the equation (1) for mc, and then you replace the values of the other parameters:
[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex] (2)
Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:
T': period of chair when the astronaut is sitting = 2.0s
M: mass of the astronaut plus mass of the chair = ?
[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)
Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :
[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]
The mass of the astronaut is 48.48 kg
An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?
Answer:2000
Explanation:
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answer:
Explanation:
a )
from lens makers formula
[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]
f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face
putting the values
[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]
1.462 = 2 - 1 / r₂
1 / r₂ = .538
r₂ = 1.86 cm .
= 18.6 mm .
b )
object distance u = 25 cm
focal length of convex lens f = 1.8 cm
image distance v = ?
lens formula
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]
[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]
.5555 - .04
= .515
v = 1.94 cm
c )
magnification = v / u
= 1.94 / 25
= .0776
size of image = .0776 x size of object
= .0776 x 10 mm
= .776 mm
It will be a real image and it will be inverted.
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)
W is described as: W = m.g
Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N
N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:
[tex]W_y[/tex] = m.g.cos(14)
Therefore, the formula will be:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
A rocket rises vertically, from rest, with an acceleration of 3.99 m/s2 until it runs out of fuel at an altitude of 775 m. After this point, its acceleration is due to gravity downwards. What is the speed of the rocket, in m/s, when it runs out of fuel?
Answer:
Vf = 78.64 m/s
Explanation:
The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:
2as = Vf² - Vi²
where,
a = acceleration = 3.99 m/s²
s = distance or height covered by rocket till fuel runs out = 775 m
Vf = Final Velocity = ?
Vi = Initial velocity = 0 m/s (Since, rocket starts from rest)
Therefore,
2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²
Vf = √(6184.5 m²/s²)
Vf = 78.64 m/s
On a brisk walk, a person burns about 331 Cal/h. If the brisk walk were done at 3.0 mi/h, how far would a person have to walk
to burn off 1 lb of body fat? (A pound of body fat stores an amount of chemical energy equivalent to 3,500 Cal.)
mi?
Answer:
32mi
Explanation:
If 1lb contains 3,500 Cal
It means the number of hours required to burn 3500cal would be;
3500/331 = 10.57hours
But a brisk walk is 3.0 mi/h,
It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles
32miles{ approximated to the nearest whole}
Note Distance = speed × time
Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.
A 20 g "bouncy ball" is dropped from a height of 1.8 m. It rebounds from the ground with 80% of the speed it had just before it hit the ground. Assume that during the bounce the ground causes a constant force on the ball for 75 ms. What is the force applied to the ball by the ground in N?
The following are not correct: 0.513 N, 0.317 N, 0.121 N. Please show your work so I can understand!
Answer:
F = 0.314 N
Explanation:
In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:
[tex]v^2=v_o^2+2gy[/tex] (1)
y: height from the ball starts its motion = 1.8 m
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
v: final velocity of the ball = ?
You replace the values of the parameters in the equation (1):
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]
Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex] (2)
m: mass of the ball = 20g = 20*10^-3 kg
v1: velocity of the ball just before it hits the ground = 5.93m/s
v2: velocity of the ball after it impacts the ground (80% of v1):
0.8(5.93m/s) = 4.75 m/s
Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s
You replace the values of the parameters in the equation (2):
[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]
The minus sign means that the force is applied against the initial direction of the motion of the ball.
The applied force by the ground on the bouncy ball is 0.314 N
write the answer:
physics ... i need help
Answer:
6 gallons
Explanation:
At 30 mph, the fuel mileage is 25 mpg.
After 5 hours, the distance traveled is:
30 mi/hr × 5 hr = 150 mi
The amount of gas used is:
150 mi × (1 gal / 25 mi) = 6 gal
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Required:
Determine the coefficient of static friction between the car and the track.
Answer:
Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)
Explanation:
The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.
Let [tex]m[/tex] represent the mass of this car.Let [tex]r[/tex] represent the radius of the circular track.This answer will approach this question in two steps:
Step one: determine the centripetal force when the car is about to skid.Step two: calculate the coefficient of static friction.For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.
Centripetal Force when the car is about to skidThe question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.
The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)
Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:
[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].
The idea is to solve for the final angular velocity using the angular analogy of that equation:
[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].
In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:
[tex]\omega(\text{initial}) = 0[/tex] since the car was initially not moving.[tex]\theta = \displaystyle \frac{\pi}{2}[/tex] since the car travelled one-quarter of the circle.Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:
[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].
Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:
[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].
Coefficient of static friction between the car and the trackSince the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].
Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.
Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:
[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].
The size of this force should be equal to that of the centripetal force when the car is about to skid:
[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].
Solve this equation for [tex]\mu_s[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].
Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].
(Three significant figures.)
A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?
Answer:
0.938 seconds
Explanation:
For the ball thrown upwards, we use the formula below to solve it:
[tex]s = ut - \frac{1}{2}gt^2[/tex]
where s = distance moved
u = initial speed = 19.2 m/s
t = time taken
g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
Let x be the height at which both balls are level, this means that:
=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)
For the ball dropped downwards, we use the formula below:
[tex]s = ut + \frac{1}{2}gt^2[/tex]
u = 0 m/s
At the point where both balls are level:
s = 18 - x
=> [tex]18 - x = 0 + 4.9t^2[/tex]
=> [tex]x = 18 - 4.9t^2[/tex]__________(2)
Equating both (1) and (2):
[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]
They will be level after 0.938 seconds
Jackson heads east at 25 km/h for 20 minutes before heading south at 45 km/h for 20 minutes. Hunter heads south at 45 km/h for 10 minutes before heading east at 40 km/h for 30 minutes. Find average velocity (magnitude and direction) of each person
Answer:
The average velocity of Jackson is 18.056 m/s South
The average velocity of Hunter is 10.65 m/s East
Explanation:
initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east
time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds
⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m
Final velocity of Jackson, v = 45 km/h South = 12.5 m/s South
time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes
time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m
Average velocity of Jackson;
[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]
initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South
time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds
⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m
Final velocity of Hunter, v = 40 km/h east = 11.11 m/s east
time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes
time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m
Average velocity of Hunter;
[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]
A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s
Answer:
a) 98 m/s
b) 19.6 s
c) 1449.8 m
d) 29.6 s
e) 168.6 m/s
f) 46.8 s
Explanation:
Given that
Acceleration of the rocket, a = 5 m/s²
Altitude of the rocket, s = 960 m
a)
Using the equation of motion
v² = u² + 2as, considering that the initial velocity, u is 0. Then
v² = 2as
v = √2as
v = √(2 * 5 * 960)
v = √9600
v = 98 m/s
b)
Using the equation of motion
S = ut + ½at², considering that initial velocity, u = 0. So that
S = ½at²
t² = 2s/a
t² = (2 * 960) / 5
t² = 1920 / 5
t² = 384
t = √384 = 19.6 s
c)
Using the equation of motion
v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that
0 = 98² + 2(-9.8) * s
9600 = 19.6s
s = 9600/19.6
s = 489.8 m
The maximum altitude now is
960 m + 489.8 m = 1449.8 m
d)
Using the equation of motion
v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that
0 = 98 +(-9.8 * t)
98 = 9.8t
t = 98/9.8
t = 10 s
Total time then is, 10 + 19.6 = 29.6 s
e) using the equation of motion
v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,
v² = 0 + 2 * 9.8 * 1449.8
v² = 28416.08
v = √28416.08
v = 168.6 m/s
f) using the equation of motion
S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s
1449.8 = 0 + ½ * 9.8 * t²
2899.6 = 9.8t²
t² = 2899.6/9.8
t² = 295.88
t = √295.88
t = 17.2 s
total time in air then is, 17.2 + 29.6 = 46.8 s
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.280 T/s.
A. What is the magnitude of the electric field induced in the ring?
B. In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1. Counterclockwise2. Clockwise
Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex] (1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex] (2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):
[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]
hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.
