Answer:
B) 12 m
Explanation:
Gravitational potential energy is:
PE = mgh
Given PE = 5997.6 J, and m = 51 kg:
5997.6 J = (51 kg) (9.8 m/s²) h
h = 12 m
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
= Mass of bean bag = 0.354 kg
= Mass of empty crate = 3.77 kg
= Speed of the bean bag
= Speed of the crate
Acceleration


From equation of motion

In this system the momentum is conserved

The speed of the bean bag is 31.42383 m/s
Answer:
(a) 3107.98 J
(b) 14530.6 J
Explanation:
mass, m = 3.56 kg
angular speed, ω = 179 rad/s
Moment of inertia of solid cylinder, I = 1/2 mr^2
where, m is the mass and r be the radius of the cylinder.
(a) radius, r = 0.330 m
I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2
The formula for the rotational kinetic energy is given by

K = 0.5 x 0.194 x 179 x 179 = 3107.98 J
(b) radius, r = 0.714 m
I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2
The formula for the rotational kinetic energy is given by

K = 0.5 x 0.907 x 179 x 179 = 14530.6 J
Answer:
The voltage is 2.114 V.
Explanation:
Given that,
Mass of both weights = 225 gm
Transducer sensitivity = 0.5 V/N
The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,
We need to calculate the net equivalent force
Using formula of force


Put the value into the formula


We need to calculate the voltage
Using formula of voltage

Put the value into the formula


Hence, The voltage is 2.114 V.
Answer:
true.
Explanation:
z-score gives an idea of how far the value of data is from the mean point. Basically is the measure of a standard deviation below or above the mean.
More the deviation of the data from the mean more is the standard deviation.
Z-score or value is calculated by the expression given below:

x is the value
μ is the mean
σ is the standard deviation
Hence, the statement given is true.