"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five t
1 answer:
Answer:
Speed of ball as it leaves ramp is 2 m/s
Explanation:
From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec
one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled
since speed of ball is given by v = thus,
the speed = 2 m/s
now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average
by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading by different students
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Answer:
the kinetic energy lost due to friction is 22.5 J
Explanation:
Given;
mass of the block, m = 0.2 kg
initial velocity of the block, u = 25 m/s
final velocity of the block, v = 20 m/s
The kinetic energy lost due to friction is calculated as;
Therefore, the kinetic energy lost due to friction is 22.5 J
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find