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Julli [10]
2 years ago
9

Amy uses 20 N of force to push a lawnmower 10 meters. How much work does she do?

Physics
1 answer:
Ira Lisetskai [31]2 years ago
8 0

Answer:

200J

Explanation:

work done = force x distance

                  = 20 x 10

                  = 200J

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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
steposvetlana [31]

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
The initial velocity of a micro van is 15 m/s. It gains a velocity of 40 ms in 10 seconds. Calculate the average velocity and ac
PSYCHO15rus [73]

Answer:

{ \bf{average \: velocity =  \frac{15 + 40}{2}}} \\   = 27.5 \:  {ms}^{ - 1}  \\  { \bf{acceleration =  \frac{v - u}{t} }} \\  =  \frac{40 - 15}{10}  \\  = 2.5 \:  {ms}^{ - 2}  \\  \\ { \tt{second \: qn : }} \\ { \bf{final \: velocity =u + at }} \\ v = 0 + (5 \times 10) \\  = 50 \:  {ms}^{ - 1}

5 0
2 years ago
Hey! anybody knows the answer then help me i will mark your answer!!!
Volgvan

10/9

Explanation:

option 2 is the correct answer

8 0
2 years ago
What is a milliliter.
zavuch27 [327]

Answer:

one thousandth of a liter (0.002 pint).

Explanation:

7 0
3 years ago
Read 2 more answers
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