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ikadub [295]
3 years ago
12

A 1200-kg car initially at rest undergoes constant acceleration for 8.8 s, reaching a speed of 10 m/ s. It then collides with a

stationary car that has a perfectly elastic spring bumper. What is the final kinetic energy of the two-car system
Physics
1 answer:
atroni [7]3 years ago
4 0

Answer:

The final kinetic energy of the two-car system is 60,000 J.

Explanation:

Given;

mass of the car, m = 1200 kg

time of motion, t = 8.8 s

final velocity of the car, v = 10 m/s

Apply the principle of conservation of kinetic energy; the initial kinetic energy is equal final kinetic energy.

K.E_i = K.E_f\\\\K.E_f = \frac{1}{2}mv^2\\\\K.E_f =  \frac{1}{2}(1200)(10)^2\\\\K.E_f = 60,000 \ J

Therefore, the final kinetic energy of the two-car system is 60,000 J.

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The key difference between the binomial and hypergeometric distribution is that
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Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

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The force F is expressed in terms of the mass “m” and acceleration “a” according to the
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Answer:

F = [M] × [L1 T-2] = M1 L1 T-2.

Explanation:

Therefore, Force is dimensionally represented as M1 L1 T-2.

5 0
3 years ago
a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s
Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
3 years ago
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
2 years ago
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