<span>2.5 m/s going upward.
In the situation described, Erica and Danny undergo a non-elastic collision which will conserve their combined momentum. Since Erica is stationary, her momentum is 0. And since Danny is moving upward at 4.7 m/s his momentum is 43 kg * 4.7 m/s = 202.1 kg*m/s. Assuming that both Erica and Danny will be moving as a joined system, their combined mass is 38 kg + 43 kg = 81 kg. Since the momentum will be the same, their velocity will be 202.1 kg*m/s / 81 kg = 2.495061728 m/s. Since we only have 2 significant figures in the provided data, rounding the result to 2 significant figures gives a velocity of 2.5 m/s going upward.</span>
Explanation and Examples
let the mass of the compressor be
mass (m):
height in x axis is (h1)
height in y axis be (h2):
Height difference: h2-h1
displacement x force:
mass x gravity x height
(m)*9.8*(height difference) = ___ J
Since gravity is forcing down, it would be negative!
Put the values that you require and get the answer.
<span>Answer: Answer is The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.</span>
Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping