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3 years ago

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If the coefficient of friction between block and surface is 0.251 , what was the speed of the bullet immediately before impact?

Answer in units of m/s.

1 answer:

lubasha [3.4K]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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If a train is going 60 m/s hits the brakes, and it takes the train 1 minute 25 seconds to stop, what is the train’s acceleration

Cloud [144]

Hey there, the answer is .............................. About 0.7 m/sec^2
<span>
Acceleration is the change in speed / time </span>

<span>Change in speed is 60 m/sec </span>

<span>Time is 1 minute 25 second. Convert that to seconds. </span>

<span>Divide the change in speed by the time in seconds.

About 0.7 m/sec^2

</span><span>So the acceleration is - 60 / 85 = - 0.71 m/s^2

HOPE I HELPED!!!!!!!!!!</span>

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2 years ago

Which of the examples below is an example of convection?

nasty-shy [4]

Basking in the sun :)

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2 years ago

How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help

Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

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2 years ago

Read 2 more answers

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

bearhunter [10]

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

Total Energy before fall = Total Energy after fall

$E_{i}$ = $E_{f}$

Here, total energy is the sum of kinetic energy and potential energy

$K.E_{i}$ + $P.E_{i}$ = $K.E_{f}$ + $P.E_{f}$ (a)

When banana is at height h, it has

$K.E_{i}$ = 0 and $P.E_{i}$ = mgh

and when it reaches the river, it has

$K.E_{f}$ = 1/2m $v^{2}$ and $P.E_{f}$ = 0

Putting the values in equation (a)

0 + mgh = 1/2m $v^{2}$ + 0

mgh = 1/2m $v^{2}$

<em>cutting 'm' from both sides</em>

<em> </em>gh = 1/2 $v^{2}$

v = $\sqrt{2gh}$

Hence, the velocity of banana before hitting the water is

v = $\sqrt{2gh}$

5 0

2 years ago

A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

AlekseyPX

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

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3 years ago