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MrMuchimi
3 years ago
13

A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

iameter is 0.190 m , to a hanging book with mass 3.05 kg . The system is released from rest, and the books are observed to move a distance 1.19 m over a time interval of 0.800 s .
A) What is the tension in the part of the cord attached to the textbook?
B) What is the tension in the part of the cord attached to the book?
C) What is the moment of inertia of the pulley about its rotation axis?
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

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Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st
pantera1 [17]

Answer:

d) v1 = v2 = v3

Explanation:

This can be answered using conservation of energy. We calculate the mechanical energy E=K+U (sum of kinetic and gravitational potential energies) at the original and final points, and impose they are equal.

At the original point we have, for the three balls:

E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh

At the final point we have, for the three balls:

E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}

Since we have E_i=E_f, and E_i is the same for all balls, then E_f is the same for all balls, which means that v_f, the final velocity, is the same for all balls.

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3 years ago
A motorcycle is moving at 18 m/s when its brakes are applied, bringing the cycle to rest in 4.7 s. To the nearest meter, how far
morpeh [17]

Answer:

the motorcycle travels 42.4 meters until it stops.

Explanation:

Vi= 18 m/s

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t= 4.7 sec

Vf= Vi - a*t

deceleration:

a= Vi/t

a= 18m/s  / 4.7 sec =>  a=-3.82 m/s²

x= Vi*t - (a * t²)/2

x= 42.4m

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2 years ago
A 25.0 kg projectile is fired by accelerating it with an electromagnetic rail gun on the earth's surface. the rail makes a 30.0
stich3 [128]

(a)

The work done on the projectile is 9375 joule.

The work on the projectile is calculated as

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=1250×7.5

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(b)

The speed of the projectile after 7.5 m is 27.38 m/s

First we need to find out the acceleration of the projectile

F=m×a

1250=25×a

a=50 m/s^{2}

Now the velocity of the projectile after 7.5 m is calculated as

v^2=u^2+2a×s

v^2=0+2×50*7.5

v=27.38 m/s

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A is the correct answer

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