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2 years ago

14

A freight train is accelerating on a level track. The tension in the coupling between the engine and the first freight car would

change if some of the cargo in the last car were transfered to other cars?

1 answer:

rewona [7]2 years ago

6 0

Answer:

False

Explanation:

According to Newton's 3rd law, the tension between the engine and the 1st freight cars would equals to the force required to pull all the freight cars from the 1st one on ward. In addition, Newton's 2nd law states that the F = ma, which means that force is proportional to the mass of all the freight cars, given that the freight train is accelerating. So if some of the cargo in the last car were transferred to other cars, their total mass would stay the same, and so as force F and tension.

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If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

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Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

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Using equation of motion

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Using equation of motion again

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The sprinter need to covers only 85 m.

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Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

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Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

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The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

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