First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!
Centripetal force = (mv^2)/r
so r = (mv^2)/ force = 246500 / 1100 = 224 m
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
Answer:
instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken