A 10 kg mass rests on a table. What acceleration will be generated when a force of 20 N is applied and encounters a frictional f
1 answer:
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Given :
Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.
A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².
To Find :
The magnitude of F.
Solution :
Torque on hoop is given by :
( Moment of Inertia of hoop is MR² )
Putting value of M, R and α in above equation, we get :
Therefore, the magnitude of force F is 25 N.
Hence, this is the required solution.
Explanation:
Formula to calculate electric field because of the plate is as follows.
E =
=
=
Now, we will consider that equilibrium of forces are present there. So,
ma = qE
a =
=
According to the third equation of motion,
or, d =
=
= 0.254 m
Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.