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Lady_Fox [76]
3 years ago
13

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle

l to the ground. The mass of the stuntman is 107 kg, and the coefficient of kinetic friction between the road and him is 0.682. Find the tension in the cable.
Physics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

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If a book has a a mass of 2 kg, how much does the book weigh?
Anestetic [448]

Answer:

A. 20 N

Explanation:

the weight of book is 2×10= 20N

7 0
3 years ago
The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50
sammy [17]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, \theta_c=52^o

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

n_1sin\theta_c=n_2sin\theta_2

Here, \theta_2=90

sin\theta_c=\dfrac{n_2}{n_1}

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

n_1=\dfrac{n_2}{sin\theta_c}

n_1=\dfrac{1}{sin(52)}

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

8 0
3 years ago
I didn't understand ​
julsineya [31]
Huh? The answer is A Shadow
4 0
3 years ago
Why does polarity allow water to be such a good solvent
Alexeev081 [22]
Water is capable of dissolving a variety of different substances, and is attracted to many other different molecules
7 0
3 years ago
A student wishes to record a 7.5-kilogram watermelon colliding with the ground. Calculate how far the watermelon must fall freel
algol [13]

Answer:

42.86m

Explanation:

The first thing we should keep in mind is that the watermelon moves with uniform acceleracion equal to gravity (9.81m / s ^ 2)

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.g} =Y

where

Vf=29m/s= final speed

Vo= initial speed=0m/S

g=gravity=9.81m/s^2

Y= distance traveled(m)

solving

\frac {Vf^{2}-Vo^2}{2.g} =Y\\\frac {(29m/s)^{2}-(0m/s)^2}{2(9.8m/s^2)} =Y\\Y=42.86m\\

the distance traveled by watermelon is 42.86m

       

5 0
3 years ago
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