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3 years ago

9

If these gas molecules were compressed further, how would the average kinetic energy of the molecules be affected? A. It would i

ncrease. B. It would decrease. C. It would change constantly. D. It would remain exactly the same

1 answer:

Advocard [28]3 years ago

8 0

I believe the correct response would be B. It would decrease.

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Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

$V2=\frac {M1v1}{m2}$

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

$V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s$

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3 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The

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Which domain bacteria involved in making sourdough bread belong to

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