Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
The answer to your question is when time = 50 s, work = 5000 J
when time = 90 s, work = 9000 J
Explanation:
Data
time = 50 s or 90 s
Power = 100 watts
Power is defined as the rate of work done per unit of time.
Power = Work / time
-Solve for Work
Work = Power x time
-Substitution
Work = 100 x 50
-Result
Work = 5000
2.-When time = 90 s
Work = 100 x 90
-Result
Work = 9000 watts
Hi
I think the answer is:
GRAVITATIONAL POTENTIAL ENERGY TRANSFORMS INTO KINETIC ENERGY.
HOPE IT HELPS.
I think the answer should be that, Kinetic energy is contained in a moving object, while potential energy exists in a stored form.
Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²