The maximum pressure variations the human ear can withstand above and below atmospheric pressure is around 30 pa. the normal atmospheric pressure is around 101325 pa. hence the variation in the maximum pressure for human ear is very small as compared to the atmospheric pressure. if the ear is exposed to a pressure greater than this , it can cause permanent damage to the ear.
The famous Newton’s Third Law states that “For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.”
By using this,
10grams or 0.01kg of bullet with speed 400 m/sec and 5kg gun recoil with speed suppose ‘v’.
0.01×400=5×v
4/5=v
v=0.8m/sec ANSWER.
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
Explanation:
1st- states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.
2nd- states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it. (most important law)
3rd- states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. (law of action/reaction)
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
