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3 years ago

Dos coches, uno de 1.000 kg que circula a 100 km/h, y otro de 1.200 kg que circula a 90

Physics

1 answer:

Kobotan [32]3 years ago

7 0

The collision of car 2 is more violent (because more impulse is exerted)

Explanation:

The collision which is more violent is the one in which more impulse is exchanged.

The impulse exerted by each car on the wall is equal to the change in momentum of the car:

$I=m\Delta v$

where

m is the mass of the car

$\Delta v$ is the change in velocity

For the car 1,

m = 1000 kg

$\Delta v = -100 km/h \cdot \frac{1000 m/km}{3600 s/h}=-27.8 m/s$ (the sign is negative because the velocity of the car has changed from 100 km/h to 0 km/h)

So the magnitude of the impulse of car 1 is

$I_1 = (1000)(27.8)=27,800 kg m/s$

For the car 2,

m = 1200 kg

$\Delta v = -90 km/h \cdot \frac{1000 m/km}{3600 s/h}=-25 m/s$

So the magnitude of the impulse of car 2 is

$I_2 = (1200)(25)=30,000 kg m/s$

So, car 2 exerts a larger impulse, therefore its impact is more violent.

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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A(n) 2602 kg van runs into the back of a(n)

forsale [732]

Answer:

8.5 m/s

Explanation:

please see paper for the work!

8 0

3 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

3 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

5 0

3 years ago

A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I

Fofino [41]

Answer:

Explanation:

Water level remains unchanged

The first thing to realize is that the buoyancy force is the same as, or equal to the weight of the wood, this same force is also the same as or equal to the weight of the water displaced by the wood. In the two cases, the weight of the wood will be unaffected nonetheless, and thus the water level will remain the same.

Therefore, the answer is B, the water level remains unchanged.

6 0

3 years ago

A coin is dropped. What is its displacement after 0.30 s.

Morgarella [4.7K]

Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.

3 0

3 years ago