A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from the ground level. Wh
1 answer:
Answer:B
Explanation:
For the ball dropped From a tall ball building the direction of acceleration is downward and its magnitude is
For the ball thrown upward towards the dropped ball the direction and magnitude of acceleration is same i.e. downwards as gravity always acting downward with constant magnitude.
Thus option B is correct
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Answer:
|v| = 8.7 cm/s
Explanation:
given:
mass m = 4 kg
spring constant k = 1 N/cm = 100 N/m
at time t = 0:
amplitude A = 0.02m
unknown: velocity v at position y = 0.01 m
1. Finding Ф from the initial conditions:
2. Finding time t at position y = 1 cm:
3. Find velocity v at time t from equation 2:
Answer:
The amount of heat required is
Explanation:
From the question we are told that
The mass of water is
The temperature of the water before drinking is
The temperature of the body is
Generally the amount of heat required to move the water from its former temperature to the body temperature is
Here is the specific heat of water with value
So
=>
Generally the no of mole of sweat present mass of water is
Here is the molar mass of sweat with value
=>
=>
Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as
Here is the latent heat of vaporization with value
=>
=>
Generally the overall amount of heat energy required is
=>
=>
Answer:
15.1°
Explanation:
The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:
Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:
(1)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the gravitational acceleration
t is the time
Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by
Substituting t into (1) we find the final vertical velocity
where the negative sign means that the velocity is downward.
Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal: