Explanation:
It is given that,
Mass of the rim of wheel, m₁ = 7 kg
Mass of one spoke, m₂ = 1.2 kg
Diameter of the wagon, d = 0.5 m
Radius of the wagon, r = 0.25 m
Let I is the the moment of inertia of the wagon wheel for rotation about its axis.
We know that the moment of inertia of the ring is given by :
The moment of inertia of the rod about one end is given by :
l = r
For 6 spokes, 
So, the net moment of inertia of the wagon is :
So, the moment of inertia of the wagon wheel for rotation about its axis is 
. Hence, this is the required solution.
Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .
Answer:
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