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Thepotemich [5.8K]

3 years ago

15

A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a m

ass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?

1 answer:

Nostrana [21]3 years ago

5 0

Answer:

Velocity of the first car after the collision, $v_1=8.93\ m/s$

Explanation:

It is given that,

Mass of the car, $m_1 = 480\ kg$

Initial speed of the car, $u_1 = 14.4\ m/s$

Mass of another car, $m_2 = 570\ kg$

Initial speed of the second car, $u_2 = 13.3\ m/s$

New speed of the second car, $v_2 = 17.9\ m/s$

Let $v_1$ is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_2v_2=m_1v_1$

$480\times 14.4+570\times 13.3-570\times 17.9=480v_1$

$v_1=8.93\ m/s$

So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.

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Mashcka [7]

Answer:

1.28 m

Explanation:

As shown in the diagram attached,

According to the principle of moment,

For a body at equilibrium,

Sum of clockwise moment = sum of anticlockwise moment.

Taking moment about the pivot,

W₁(1.6)+W(0.133) = W₂(x)............... Equation 1

Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.

But,

W = mg

Where g = 9.8 m/s², m = mass of the body

Therefore,

W₁ = 26×9.8 = 254.8 N,

Wₓ = 18×9.8 = 176.4 N

W₂ = 34.4×9.8 = 337.12 N

Substitute these values into equation 1

(254.8×1.6)+(176.4×0.133) = 337.12(x)

407.68+23.4612 = 337.12x

337.12x = 431.1412

x = 431.1412/337.12

x = 1.2789

x ≈ 1.28 m

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2 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

matrenka [14]

Answer:

a) 6.1 m

b) 4.6 s

c) 1.326 m/s

d) 0.325 m

Explanation:

a) The wave length is the distance between 2 crests λ = 6.1m

b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s

c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s

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An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of p

FromTheMoon [43]

The given question is incomplete. The complete question is as follows.

An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.

Explanation:

It is known that Boltzmann formula of entropy is as follows.

s = k ln W

where, k = Boltzmann constant

W = number of energetically equivalent possible microstates or configuration of the system

In the given case, W = 36. Now, we will put the given values into the above formula as follows.

s = k ln W

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Thus, we can conclude that the entropy of this system is $4.945 \times 10^{-23} J/K$ .

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$v=u+gt$

where

v = 0 is the velocity at the maximum height

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Solving for t,

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The total time of the motion is twice this value, so:

$t=2\frac{v-u}{g}$

So we see that it is inversely proportional to g.

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2 years ago