Answer:
5760 J
Explanation:
From the question given above, the following data were obtained:
Mass of block = 48 kg
Height (h) = 12 m
Gravitational field strength (g) = 10 N/Kg
Gravitational potential energy (PE) =?
The gravitational potential energy stored by the block can simply be obtained as follow:
PE = mgh
PE = 48 × 10 × 12
PE = 5760 J
Therefore, the gravitational potential energy stored by the block is 5760 J
Answer: Technician A is correct
Explanation:
The intake manifold is the compactment that all fuel and air supply to the cylinders. It's connected to the engine so it has to be disconnected while the exhaust manifold receives all the exhaust gases from the cylinders and releases the gas through a single or double exhaust gases outlet.
Answer:
0.72 Hz minimum frequency
Explanation:
When the damping is negligible,Amplitude is given as

here
= (6.30)/(0.135) = 46.67 N/m kg
= 1.70/(0.135)(0.480) = 26.2 N/m kg
From the above equation , rearranging for ω,

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47
⇒ ω = 8.53 or 4.52 rad/s
Frequency = f
ω=2 π f
⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz
The lower frequency is 0.72 Hz and higher is 1.36 Hz
Answer:
The amount of work done required to stretch spring by additional 4 cm is 64 J.
Explanation:
The energy used for stretching spring is given by the relation :
.......(1)
Here k is spring constant and x is the displacement of spring from its equilibrium position.
For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

k = 4 x 10⁴ N/m
Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).
Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = 72 J
But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :
E = ( 72 - 8 ) J = 64 J