Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction


Put the value into the formula


Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Answer:

Explanation:
Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

The era after the KT event occurred
The engine's efficiency is (35J)/125J) = 28% .
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Answer:
The horizontal component of the velocity is 21.9 m/s.
Explanation:
Please see the attached figure for a better understanding of the problem.
Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.
To find vx, let´s use the following trigonometric rule of right triangles:
cos α = adjacent / hypotenuse
cos 5.7° = vx / 22 m/s
22 m/s · cos 5.7° = vx
vx = 21.9 m/s
The horizontal component of the velocity is 21.9 m/s.