Answer:
675J
Explanation:
Given parameters:
Force = 45N
Distance = 15m
Unknown:
Work done by Sheila = ?
Solution:
Work done by a body is the amount of force applied to make a body move through a distance;
Work done = Force x distance
Now;
Work done = 45 x 15 = 675J
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
In this problem we have the electric field intensity E:
E = 6.5 ×
newtons/coulomb
We have the magnitude of the load:
q = 6.4 ×
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 ×
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 ×
)(6.5 ×
)(1.2 ×
)
PE = 5.0 x
joules
None of the options shown is correct.