Answer:
b. 600,000 J
Explanation:
Applying the law of conservation of energy,
The thermal energy created = Kinetic energy of the suv.
Q' = 1/2(mv²)............... Equation 1
Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.
From the question,
Given: m = 3000 kg, v = 20 m/s
Substitute these values into equation 1
Q' = 1/2(3000×20²)
Q' = 600000 J
Hence the right option is b. 600,000 J
The energy carried by the incident light is
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
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>
Use the Pythagoras for the magnitude and the tan^-1 x = -1 for the angle
displacement = 4^2 + 4^2 = 32 = 4 sqrt(2) = 5.65 km
angle is 135 degrees.
