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Artyom0805 [142]

3 years ago

5

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

the train is initially moving at a speed of 49.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

1 answer:

mart [117]3 years ago

5 0

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

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3 years ago

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

guapka [62]

<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=200m/s$ is the bullet's initial speed

$\theta=0$ because we are told the bullet is shot horizontally

$t$ is the time since the bullet is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the bullet

$y=0$ is the final height of the bullet (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

<h2>Part (a):</h2>

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

$0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}$ (3)

$0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}$ (4)

Finding $t$ :

$t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}$ (5)

Then we have the time elapsed before the bullet hits the ground:

$t=0.553s$ (6)

<h2>Part (b):</h2>

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

$x=V_{o}cos\theta t$ (1)

Substituting the knonw values and the value of $t$ found in (6):

$x=200m/s.cos(0)(0.553s)$ (7)

$x=200m/s(0.553s)$ (8)

Finally:

$x=110.656m$

4 0

3 years ago

What is center of mass?

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3 0

3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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3 years ago