A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V
1 answer:
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When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
(2)
where
is the refractive index of the second medium and 
is the refractive index of the first medium.
We can find the ratio
by using Snell's law:
(1)
where
is the angle of incidence
is the angle of refraction
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
where
We can find the ratio
where
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
Answer:
The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.
Explanation:
Given:
Radius of the curve (R) = 150 m
Maximum speed of truck (v) = 35.5 m/s
Let the coefficient of friction between the roadway and the wheels of the truck be "μ".
As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.
The centripetal force acting on the truck is given as:
Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.
Frictional force is given as:
Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,
Therefore, frictional force,
Now, frictional force = centripetal force
Plug in the given values and solve for 'μ'. This gives,
Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833