- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s
- Velocity of mobile after 45 seconds
We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.
We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
Explanation:
<u>Elastic Potential Energy
</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:
Substituting:
Calculating:
Answer: 2.068*
m
Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.
workdone= qV
energy = hc/λ
q=magnitude of an electronic charge= 1.602*
h= planck constant = 6.626*
c= speed of light =2.998* 
v= potential difference= 6*
λ= wavelength=unknown
by making λ subject of formulae we have that
λ= 
λ = 6.626* * 2.998* / 1.602* * 6*
λ = 
by doing the necessary calculations, we have that
λ = 2.068*m
Answer:
<h3>4</h3>
Explanation:
VR = Velocity ratio = Distance moved by effort/Distance moved by load
Given:
Distance moved by effort = 20m
Distance moved by Load = 5m
VR = 20/5
VR = 4
Hence the velocity ratio of the car is 4
