As the ball is moving in air as well as we have to neglect the friction force on it
So we can say that ball is having only one force on it that is gravitational force
So the force on the ball must have to be represented by gravitational force and that must be vertically downwards
So the correct FBD will contain only one force and that force must be vertically downwards
So here correct answer must be
<em>Diagram A shows a box with a downward arrow. </em>
Answer:
It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.
Explanation:
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:
option (B)
Explanation:
Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.
Its unit is N/m².
The formula for the Young's modulus is given by

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.
It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.