A baseball is struck home plate and acquires a speed of 60m/s. It rises to a height of 100.0 m above its starting level. Find it
1 answer:
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Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
Subsequently,
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 .
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
Answer:
horizontal component of normal force is equal to the centripetal force on the car
Explanation:
As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle
This force is due to friction force when car is moving in circle with uniform speed
Now it is given that car is moving on the ice surface such that the friction force is zero now
so here we can say that centripetal force is due to component of the normal force which is due to banked road
Now we have
so we have
so this is horizontal component of normal force is equal to the centripetal force on the car
Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
But heat capacity of object B is twice that of object A
Therefore, the final temperature of both objects is 400 K.