Answer:
5.88×10⁸ W
Explanation:
Power = change in energy / time
P = mgh / t
P = (m/t) gh
P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)
P = 5.88×10⁸ W
Answer:
<h3>The answer is 500 kg</h3>
Explanation:
The mass of the object can be found by using the formula
v is the velocity
KE is the kinetic energy
From the question we have
We have the final answer as
<h3>500 kg</h3>
Hope this helps you
Answer:
The electron’s velocity is 0.9999 c m/s.
Explanation:
Given that,
Rest mass energy of muon = 105.7 MeV
We know the rest mass of electron = 0.511 Mev
We need to calculate the value of γ
Using formula of energy
Put the value into the formula
We need to calculate the electron’s velocity
Using formula of velocity
Put the value into the formula
Hence, The electron’s velocity is 0.9999 c m/s.
Answer:
correct me if i'm wrong nut i thinks its a The atomic symbol of an element surrounded by valence electrons
Explanation:
The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.
<h3>What is velocity?</h3>
Velocity can be defined as the ratio of the displacement and time of a body.
To calculate the final velocity of Skater 1 we use the formula below.
Formula:
- mu+MU = mv+MV............ Equation 1
Where:
- m = mass of the first skater
- M = mass of the second skater
- u = initial velocity of the first skater
- U = initial velocity of the second skater
- v = final velocity of the first skater
- V = final velocity of the second skater.
make v the subject of the equation.
- v = (mu+MU-MV)/m................ Equation 2
Note: Let left direction represent negative and right direction represent positive.
From the question,
Given:
- m = 105 kg
- u = -2 m/s
- M = 71 kg
- U = 5 m/s
- V = -3.4 m/s.
Substitute these values into equation 2
- v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
- v = (-210+355+241.4)/105
- v = 386.4/105
- v = 3.68 m/s
- v ≈ 3.7 m/s
Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.
Learn more about velocity here: brainly.com/question/25749514
