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marissa [1.9K]
2 years ago
15

Stunt car A and stunt car B are identical cars with the same mass of 35.9 kg. They are both traveling at 37.1 m/s. Stunt car A c

rashes into a hard concrete wall, whereas stunt car B crashes into a big soft mattress. 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:
frez [133]2 years ago
8 0

Answer:

<h2>Stunt car A and stunt car B are identical cars with the same mass of 35.9 kg. They are both traveling at 37.1 m/s. Stunt car A crashes into a hard concrete wall, whereas stunt car B crashes into a big soft mattress. They both come to a complete stop after the impact.Stunt Car A experiences a <u>LARGE FORCE</u> over a <u>SHORT PERIOD</u> of time. Stunt Car B experiences a <u>SMALL FORCE</u> over a <u>LONG PERIOD</u> of time. Because of the force experienced by Stunt Car A, it will sustain <u>MORE</u> damage than Stunt Car B.</h2>

Explanation:

As we know that force is given as rate of change in momentum

so it is defined as

F = \frac{dP}{dt}

so here we know that both the cars are identical and both are moving with same speed so after coming to rest both the cars have same change in momentum

while Car A will experience more force because due to collision with concrete wall the time of collision is very small as compared to Car B

So Force on Car A is more than the force on Car B

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Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

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What do submarines translate into pictures?
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It lets the viewer know it's something to do with underwater.
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3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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defon
225 = 1/2 (50) (v2)
225 = 25 (v2)
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√9 = v
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If you went to a planet with an acceleration due to gravity of 20 m/s^2 and twirled a bucket of water, what happens to the angul
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