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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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A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 17.3 m/s directly upward.
elena-s [515]

Answer:

h=15.27m

Explanation:

Since at maximum height the vertical velocity must be null it's better to use the formula:

v_f^2=v_i^2+2ad

We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:

0=v_i^2+2ah

or

h=-\frac{v_i^2}{2a}

which for our values is:

h=-\frac{(17.3m/s)^2}{2(-9.8m/s^2)}=15.27m

7 0
3 years ago
a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis
KatRina [158]
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m +  201.67 m = 465.67 or  approximately 4.7×10² m.
7 0
3 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
3 years ago
What is the term for the force exerted by an object when it is pushed by another object?
kolezko [41]
Action-reaction pairs.

This is in reference to Newton’s second law of motion.
5 0
3 years ago
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A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a
vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be  

v_ox=v_o*cosФ

       =60*cos (30)

      = 51.96 m/s^-1

3 0
3 years ago
Read 2 more answers
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