Answer:
This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time
Explanation:
I hope this is what you need
PLEASE MAKE ME BRAINLIEST
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2
Find:
Horizontal displacement
Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s
Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
Visceral epithelial cells
Answer:
The only work done is when the person lifts the sack over a distance, W = 78.48 [N]
Explanation:
We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.
W = F * d
where:
F = force [N]
d = distance [m]
The force is given by the producto of the mass by the gravity.
F = 5 * 9.81 = 49.05 [N]
W = 49.05 * 1.6 = 78.48 [N]
