Not yet answered Points out of 2.00 Not flaggedFlag question Question text Two cars collide head-on on a level friction-free roa
1 answer:
Answer:
Velocity of center of mass must be ZERO
Explanation:
As we know that two cars are moving on friction-less surface
So here we can say that on the system of cars there is no external force along x direction
So we can say that
so here if net force is zero
since finally the two cars comes to rest so here final velocity of center of mass of two cars will be zero and hence it will be always zero as there is no force on the system of two cars
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<h2>Hello!</h2>
The answer is:
The first option, the walker traveled 360m more than the actual distance between the start and the end points.
Why?
Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).
So, calculating we have:
Traveler:
Actual distance between the start and the end point (displacement):
Now, to calculate how much farter did the traveler walk, we need to use the following equation:
Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.
Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.
Have a nice day!
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- = 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane