The coefficient of friction between the Tyre and the ground is 0.11
<u>Explanation:</u>
Given:
Radius of the track (r)=125 m.
Speed with which the car travels (v) =42 km/hr
=11.67 m/s
To Find:
Coefficient of friction between the Tyre and the ground.
Formula to be used:
We know that,Frictional force is equal to centripetal force
Frictional force=μmg
therefore 1.08 m=μmg
Cancelling "m" on both sides we get,
μ=1.08/g=1.08/9.8
=0.11
Thus the coefficient of friction between the Tyre and the ground is 0.11
True the only way to emit ur proton is if there are ground electrical
Answer:
When oil and water are combined, they do not mix evenly, but instead form two separate layers.
Explanation:
By definition, a pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.
Explanation:
Let
= distance traveled while accelerating
= distance traveled while decelerating
The distance traveled while accelerating is given by
We need the velocity of the rocket after 30 seconds and we can calculate it as follows:
This will be the initial velocity when start calculating for the distance it traveled while decelerating.
Solving for we get
Therefore, the total distance x is
Answer:
= 3.36 mm
Explanation:
From Ohm's law,
(Voltage = Current * Resistance)
The geometric definition of resistance is
where is the resistivity of the material, and are the length and cross-sectional area, respectively.
Since the wire is assumed to have a circular cross-section, its area is given by
where is the diameter.
Resistivity of copper = . With these and other given values,