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LuckyWell [14K]
3 years ago
7

Complete the sentences below correctly. Stunt Car A and Stunt Car B are identical cars with the same mass of 48.9 kg. They are b

oth traveling at 82.5 m/s. Stunt Car A crashes into a brick wall, while Stunt Car B crashes into a big pile of leaves. They both come to a complete stop after the impact. Stunt Car A experiences a _____________________ over a _____________ of time. Stunt Car B experiences a _____________________ over a _____________ of time. Because of the force experienced by Stunt Car A, it will sustain ____________ damage than Stunt Car B.
The possible answers are: short period, large force, more, long period, small force, less.
Physics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

Stunt Car A experiences a large force over a short period of time. Stunt Car B experiences a small force over a long period of time. Because of the force experienced by Stunt Car A, it will sustain more damage than Stunt Car B.

Explanation:

Both cars have the same mass and velocity, therefore they have the same momentum. During the collision, the total momentum of the car A and brick wall is conserved as well as the total momentum of the car B and the pile of leaves.

However, if we are to investigate the damage on each car, we should look at the cars not the whole system. So, the momentum difference between the cars gives us the impulse that the car felt.

\vec{J} = \Delta \vec{P}\\\vec{F}\Delta t = \vec{P}_2 - \vec{P}_1

Since the Car A will crash the wall quicker than the other car crashes through the pile of leaves,

\Delta t_1 < \Delta t_2

which gives us

F_1 > F_2

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In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri
Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

E = \frac{Q}{4\pi \varepsilon_o r^2}

where;

  • E is the electric field
  • Q is the charge
  • r is the radius

The electric field reduces by a factor of \frac{1}{r^2}

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

A = 4\pi r^2

<h3>Change in area with r</h3>

\frac{dA}{dr} = 8\pi r

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0
2 years ago
Any of two or more forms of the same element which differ only in the number of neutrons their atoms contain is called an
telo118 [61]
C. isotope

Isotopes have the same atomic number but different mass number because of the difference in the number of neutrons in the nucleus of the atom.
8 0
3 years ago
A soccer ball of mass 0.35 kg is rolling with velocity 0, 0, 1.8 m/s, when you kick it. Your kick delivers an impulse of magnitu
Mademuasel [1]

Answer:

position 9.58 m

Explanation:

In impulse exercises and amount of movement, we always assume that the contact time is small,

     I = Δp

With this expression we can calculate the final speed

     I = m Vf - m Vo

    Vf = (I + mVo) / m

    Vf = (1.8 + 0.35 1.8) /0.35

    Vf = 6.94 m / s

To calculate the acceleration of the ball we use Newton's second law, after finishing the impulse

     ∑ F = m a

     fr = m a

     a = fr / m

     a = -0.26 / 0.35

     a = -0.74 m/s²

A negative sign indicates that this acceleration is slowing the ball

Now we have speed and time acceleration, so we can use the kinematic equations to find the position at 1.5 s

   

      X = Vo t + ½ to t²

 In this case Vo is the speed with which the ball comes out after the impulse 6.94

     X = 6.94 1.5 + ½ (-0.74) 1.522

     X = 9.58 m

7 0
3 years ago
The tape in a videotape cassette has a total
xxMikexx [17]

Answer:

w=19.76 \ rad/s

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

w=w_o+\alpha \ t

Where \alpha is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

\displaystyle v_t=w.r

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call w_1 the angular speed of the loaded reel and w_2 that from the empty reel. We have

w_1=w_{o1}+\alpha_1 \ t

w_2=w_{o2}+\alpha_2 \ t

If r_f=35\ mm=0.035\ m is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

\displaystyle w_{01}=\frac{v_t}{r_f}

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

t=1.8\ h=1.8*3600=6480\ sec

\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s

\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s

If r_e=10\ mm=0.001\ m is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s

The full reel goes from w_{01} to w_{02} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}

\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2

The empty reel goes from w_{02} to w_{01} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2

So the equations for both reels are

w_1=1.098+0.00576 \ t

w_2=38.426-0.00576 \ t

They will be the same when

1.098+0.00576 \ t=38.426-0.00576 \ t

Solving for t

\displaystyle t=\frac{38.426-1.098}{0.0115}

t=3240 \ sec

The common angular speed is

w_1=1.098+0.00576 \ 3240=19.76 \ rad/s

w_2=38.426-0.00576 \ 3240=19.76 \ rad/s

They both result in the same, as expected

\boxed{w=19.76 \ rad/s}

6 0
4 years ago
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