If you found some sandstone in an ancient river bed, which of these conditions at the time that the sandstone was formed would b
1 answer:
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Answer:
Work done in lifting to a height of 1.5 m is 42.63 J.
Work done in carrying it to 50 m is 0 J
Explanation:
Given:
Mass of the pumpkin (m) = 2.9 kg
Vertical displacement of the pumpkin (y) = 1.5 m
Horizontal displacement of the pumpkin (x) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Work done by a force is given by the formula:
Where,
- Now, the work done is maximum when both force and displacement is in same direction. (
)
- Work done is minimum when both force and displacement are in opposite direction. (
)
- Work done is zero when force and displacement are perpendicular to each other. (
)
Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.
So, work done is given as:
Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.
So, Force = mg =
So, work,
So, work done in lifting the pumpkin is 42.63 J.
Now, when the pumpkin is carried to the check-out stand, the displacement is horizontal while the force applied is still in the upward direction.
So, the force and displacement are perpendicular to each other. Thus, the work done is zero.
Answer:
The answer is below
Explanation:
Firstly, the frequency is received by the wall and then it is reflected and received by the bat.
The frequency received by the wall (f') is given by:
f' =
The - sign is used when the observer is moving away from the source and + sign when the observer is moving towards the source.
Since the bat is moving towards the wall, we use a positive sign. Hence:
f' =
The frequency reflected and received by the bat f" is:
f'' =
- sign is used when the source moves toward the observer and + is used when the source moves away
since the bat moves towards the wall, then::
f'' =
v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore: