The correct answer to the question is D). Kinetic to electrical.
EXPLANATION:
Hydro electric power plants are the electricity generation plants which capture the energy of falling water to produce current.
The water present at the top of a dam has potential energy. When the water is allowed to fall from certain height, the potential energy of the water is converted into kinetic energy.
The falling water moving with high speed is allowed to fall on a turbine. The kinetic energy of the water will rotate the turbine. The turbine is attached to a generator which will produce electricity due to the electromagnetic induction.
Hence, we see that the kinetic energy of the water is converted into electric energy.
Answer:
1:04-1:10 hours
Explanation:
You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:
- In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
- You'll check and keep the letter in that 47 row (S) for future use.
- Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
- Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.
I hope you find this information useful and interesting! Good luck!
Answer:β^+
Explanation: it is beta + , also positron emission radiation
To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

Where,
Angular Displacement
Angular Acceleration
Angular velocity
Initial angular displacement
For this case we have neither angular velocity nor initial angular displacement, then

Re-arrange for 

Replacing our values,


Therefore the ANgular acceleration of the mass is 
Answer:

Explanation:
R = Horizontal range of projectile = 75 m
v = Velocity of projectile = 37 m/s
g = Acceleration due to gravity = 
Horizontal range is given by

The angle at which the arrow is to be released is
.