Examples of devices that convert electrical energy into mechanical energy — in other words, devices that use electrical energy to move something — include:
the motor in today’s standard power drills
the motor in today’s standard power saws
the motor in an electric tooth brushes
the engine of an electric car
the motor in a fan
the motor in a remote control cars that runs on batteries. So your answer would be B. Motorsport
-- Whether an object floats or sinks in water depends on whether its density is more or less than the density of water.
-- The density of the rock didn't change when she threw it into the lake.
-- The density of the water didn't change when she threw the rock into the lake.
-- The density of anything never depends on how much of it you have.
-- If the rock sank in two ounces of water, it'll sink in a billion gallons of water.
-- Choose ' <em>A </em>' .
Answer : Option D) A dropped wallet is kicked around the floor of a busy train station.
Explanation : The description that best suits the model for the energy transfer that occurs in the radiative zone is -
<h3>A dropped wallet is kicked around the floor of a busy train station.</h3>
As in the radiation zone, which is also called as radiative zone or radiative region in the layer of a star's interior where energy is primarily transported toward the exterior by means of radiative diffusion and thermal conduction. This can be correlated with the example of a dropped wallet in a busy train station as the wallet would be kicked by many people who are in a hurry to travel, this represents the radiative diffusion. So, every time the wallet is being kicked by someone the energy is getting transformed.
Answer:
The answer is a2 = 4.48 m/s2
Explanation:
Let´s start with the equations we know:
Where:
=> Final velocity
=> Initial velocity- a => acceleration
- t => time
Now, let´s divide the problem in Stage 1 and 2 and get equations for each stage:
Stage 1 knowns and unknowns:
Stage 1 equation:
Stage 2 knowns and unknowns:
Stage 2 equation:
Now we can substitute the resultant equation from stage 1 into stage´s 2 equation:
We can see "t" is on both sides, so it cancels out and we are left with:
![a_{2} = 4.48 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=a_%7B2%7D%20%3D%204.48%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
Answer:
The maximum height reached in the second trial is 16times the maximum height reached in the first trial.
Explanation:
The following data were obtained from the question:
First trial
Initial speed (u) = v
Final speed (v) = 0
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Next, we shall obtain the expression for the maximum height reached in each case.
This is illustrated below:
First trial:
Initial speed (u) = v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₁) =.?
v² = u² – 2gh₁ (going against gravity)
0 = (v)² – 2 × 9.8 × h₁
0 = v² – 19.6 × h₁
Rearrange
19.6 × h₁ = v²
Divide both side by 19.6
h₁ = v²/19.6
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₂) =.?
v² = u² – 2gh₂ (going against gravity)
0 = (4v)² – 2 × 9.8 × h₂
0 = 16v² – 19.6 × h₂
Rearrange
19.6 × h₂ =16v²
Divide both side by 19.6
h₂ = 16v²/19.6
Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.
This is illustrated below:
Second trial:
h₂ = 16v²/19.6
First trial:
h₁ = v²/19.6
Second trial : First trial
h₂ : h₁
h₂ / h₁ = 16v²/19.6 ÷ v²/19.6
h₂ / h₁ = 16v²/19.6 × 19.6/v²
h₂ / h₁ = 16
h₂ = 16 × h₁
From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.
