The gravitational potential energy of the brick is 25.6 J
Explanation:
The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.
Near the surface of a planet, the gravitational potential energy is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the brick in this problem, we have:
m = 8 kg is its mass
g = 1.6 N/kg is the strenght of the gravitational field on the moon
h = 2 m is the height above the ground
Substituting, we find:
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Models show how the atoms in a compound are connected.
Answer:
Part A - 3N/m
Part B - see attachment
Part C - 4.9 × 10-³J
Part D - E = 1/2kd² + 1/2mv² + mgh
Explanation:
This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.
The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).
The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².
Part D
Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential
energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.
The relationship between the frequency and wavelength of a wave is given by the equation:
v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency.
If we divide the equation by f we get:
λ=v/f
From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases.
So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
