Answer:
2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）
was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:
Calculate the works performed by the force first.)
I figured that is 8.2m/S，I am just not sure can anyone help me i much appreciate it.
It’s E we just had a test in this and I got it right
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
Answer:
Charge of particle 2, 
Explanation:
Given that,
Charge 1, 
The distance between charges, r = 0.241 m
Force experienced by particle 1, F = 3.44 N
We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :
or
So, the magnitude of electric charge 2 is . Since, the force is attractive then the magnitude of charge 2 must be negative.
Answer:
The magnitude of the average emf induced in the coil is approximately o.32 V.
Explanation:
As, no. of loops = N = 10
Area = 0.23 m^2
Magnetic field = B = 0.047 T
Δt = 0.34 s
These are the given things now we have to find the EMF. So the formula of emf is:
emf = e = N×ΔФ/Δt
Therefore, ΔФ = B . ΔA = BΔAcos(θ)
Since, flux is maximum so θ = 0 and cos(0) = 1
Then,
ΔФ = BA
e = N×(BΔA)/Δt
e = 10×(0.047×0.23)/0.34
e = 0.317 V.
So, we can write it by rounding off the digit
e ≅ 0.32 V.
