Answer:
Explanation:
Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s
Radio waves have the lowest radiant energy, hope this helps :]
<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.
Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.
So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.
The forces become 1/16 of their original magnitude.<span>
</span>
Answer:
Part a)

Part b)

Part c)

Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have



So it will read same as that of the mass

Part b)
When elevator is decending with constant speed then we will have



So it will read same as that of the mass

Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have



Reading is given as



Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass