Answer:
2.55 m/s
Explanation:
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.
Solution:
The work done by friction is given as:
The work done by gravity is:
Answer:
Explanation:
Work is the product of force and distance.
We know the force to lift the rock was 50 Newtons. The work was 150 Joules.
- 1 Joule is equal to 1 Newton meters.
- We can convert the units to make the problem simpler later. The work is also 150 Newton meters.
Substitute the values into the formula.
We want to solve for distance, so we must isolate the variable. Divide both sides of the equation by 50 Newtons.
The Newtons will cancel out.
The rock was lifted <u>3 meters.</u>
Answer:
Double helix
Explanation:
The Double helix is a DNA molecule. The two strands around the Double Helix is called the twisted ladder.
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
