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OLga [1]
3 years ago
8

Which of the following is not an example of an engine?

Physics
1 answer:
SashulF [63]3 years ago
4 0
The answer is A. waterfall
To be considered as an engine , it should be a Man-made objects that could be used to produce power that creates motions.
From all the options above could be used to produce such power, but the waterfall is not made by mandkind
You might be interested in
A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at
yan [13]

Answer:

a) s_{T} = 30\,m, b) t = 5\,min, c) \Delta t = 6.667\,s, d) \Delta s_{R} = 33.333\,m, e) t' = 11.667\,s, f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

s_{T} = v_{T}\cdot t

Where:

t - Time, measured in seconds.

v_{T} - Velocity of the turtle, measured in meters per second.

s_{T} - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)

s_{T} = 30\,m

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

t = \frac{s_{T}}{v_{T}}

t = \frac{150\,m}{0.5\,\frac{m}{s} }

t = 300\,s

t = 5\,min

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

v_{R} = v_{o,R} + a_{R}\cdot \Delta t

Where:

v_{R} - Final velocity of the rabbit, measured in meters per second.

v_{o,R} - Initial velocity of the rabbit, measured in meters per second.

a_{R} - Acceleration of the rabbit, measured in \frac{m}{s^{2}}.

\Delta t - Running time, measured in second.

\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}

\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }

\Delta t = 6.667\,s

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

Where \Delta s_{R} is the travelled distance of the rabbit from rest to maximum speed.

\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}

\Delta s_{R} = 33.333\,m

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

t' = \frac{\Delta s_{R}}{v_{R}}

t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }

t' = 11.667\,s

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

t_{T} = 300\,s

Rabbit:

t_{R} = 60\,s + 6.667\,s + 11.667\,s

t_{R} = 78.334\,s

The rabbit won the race as t_{R} < t_{T}.

7 0
3 years ago
12. A measuring cylinder has 150ml. of water in it. When a stone is immersed into the measuring cylinder, the water level raised
skelet666 [1.2K]

Answer:

60ml

Explanation:

I'm going to assume you mean 210ml not centimeters. To find the volume all we do is subtract both values or with the formula [ f - i = v ]  where f = final amount and i = initial amount.

210 - 150 = 60ml

Best of Luck!

4 0
3 years ago
Read 2 more answers
The acceleration of the object may be zero and the velocity of the object may not be equal to zero correct or not
Zinaida [17]

Answer:

Acceleration is the change in velocity, not the velocity itself; therefore, an object can have zero velocity but not zero acceleration because the velocity will be changing.

8 0
3 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
torisob [31]

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

8 0
3 years ago
A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h
katen-ka-za [31]

Answer:

v_r=5.89\ m.s^{-1}

Explanation:

Given:

  • mass of rocket, m_r=50\ g
  • time of observation, t=2\ s
  • mass lost by the rocket by expulsion of air, m_a=10\%\ of m_r=5\ g
  • velocity of air, v_a=53\ m.s^{-1}

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

m_a.v_a=(m_r-m_a)\times v_r

5\times 53=(50-5)\times v_r

v_r=5.89\ m.s^{-1}

7 0
3 years ago
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