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:
a
b
c
Explanation:
From the question we are told that
The angle of incidence is
The refractive index of water is
Generally Snell's law is mathematically represented as
Here is the refractive index of air with value
is the angle of refraction
So
=>
=>
Given that the angle should not be greater than then the angle of incidence will be
=>
=>
Generally for critical angle is mathematically represented as
=>
=>
Answer:
0.65 kg*m/s and 0.165 kg*m/s
Explanation:
Step one:
given data
mass m= 0.5kg
initial velolcity u=1.3m/s
final velocity v= 0.97m/s
Required
The change in momentum
Step two:
We know that the expression for impulse is given as
Ft= mv
Ft= 0.5*1.3
Ft= 0.65 kg*m/s
The expression for the change in momentum is given as
P= mΔv
substitute
Pt= 0.5*(1.3-0.97)
Pt= 0.5*0.33
Pt=0.165 kg*m/s