Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :
F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
Answer:
Explanation:
A )
The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³
B )
Tension in the ball will be equal to net force acting on the ball
Net force on the ball = buoyant force - weight .
4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )
= 40.65 x 10⁻⁶ N .
C )Tension in the 3 rd ball will be equal to net force acting on the ball
Net force on the ball = weight - buoyant force
= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )
= 121.6 x 10⁻⁶ N .
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>
Answer:
Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
Explanation:
The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):
José
Txomin
Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
