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NeTakaya
2 years ago
9

Find the acceleration of a box of mass 3.0 kg, if applied force is 20.0 N at 37° above the x axis, and the coefficient of kineti

c friction is 0.20.​
Physics
1 answer:
ExtremeBDS [4]2 years ago
4 0

Hello:

First, <u>the normal force</u> in y axis:

∑Fy = 0

∑Fy = N - Fy

∑Fy = N - 20 N * sen 37°

∑Fy = N = 12,87 N

Formula:

R = ma

F - F' = ma

F - (μN) = ma

20 N - (0.2 * 12,87 N) = 3 kg * a

20 N - 2.574 N = 3 kg * a

17,426 N / 3 kg = a

a = 5,8 m/s²

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olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

Angle=3^{\circ}

as the arrow travels it acquire a vertical velocity v_y

v_y=u+at

v_y=0+9.81\times t-------1

Range is given by

R=ut

where u=initial velocity

68=u\times t

t=\frac{68}{u}

substitute the value of t in eqn 1

v_y=9.81\times \frac{68}{u}

v_y\times u=9.81\times 68=667.08--------2

and tan(3)=\frac{v_y}{u}

v_y=utan(3)=0.0524u

substitute it in 2

0.0524 u^2=667.08

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u=112.82 m/s

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3 years ago
Star temperature is indicated by?
astraxan [27]
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3 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

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r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

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Answer:

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How much force is applied if a 130kg mass is accelerated at 5 m/s^2?​
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Answer:

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