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:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So, (Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:
Plug in the given values and solve for 'u'. This gives,
Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.
Answer:
62.06 g/mol
Explanation:
We are given that a solution containing 10 g of an unknown liquid and 90 g
Given mass of solute ==10 g
Given mass of solvent==90 g
Freezing point of solution =-3.33C
Freezing point of solvent =C
Change in freezing point =Depression in freezing point
=Freezing point of solvent - freezing point of solution=0+3.33=
Hence, molar mass of unknown liquid is 62.06g/mol.