Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Answer:
Interchanging the poles of the magnet
Reversing the direction of the applied current
Explanation:
- The working of the electric motor is associated with Fleming's left-hand rule.
- It states that if a current-carrying conductor is placed inside a magnetic field, it experiences a force in the direction perpendicular to the direction of the electric current and magnetic field.
- These three physical quantities are placed in a mutually perpendicular direction.
- So, in order to reverse the direction of force, you have to either change the direction of the current or magnetic field.
<h2>
Option 2 is the correct answer.</h2>
Explanation:
Elastic collision means kinetic energy and momentum are conserved.
Let the mass of object be m and M.
Initial velocity object 1 be u₁, object 2 be u₂
Final velocity object 1 be v₁, object 2 be v₂
Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s
Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M
Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J
Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M
We have
Initial momentum = Final momentum
24 = 3v₁ + 6M
v₁ + 2M = 8
v₁ = 8 - 2M
Initial kinetic energy = Final kinetic energy
96 = 1.5 v₁² + 18 M
v₁² + 12 M = 64
Substituting v₁ = 8 - 2M
(8 - 2M)² + 12 M = 64
64 - 32M + 4M² + 12 M = 64
4M² = 20 M
M = 5 kg
Option 2 is the correct answer.
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:
