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iris [78.8K]
3 years ago
9

John is gardening and finds a tree root in the soil. He decides he will try to pull it out. He grabs onto it and uses a force of

100 Newtons. He exerts that force for 5 seconds until he is able to yank the root out, 1 meter away. How much power did John use to extract the root?
20 joules/sec
25 joules/sec please help due soon
15 joules/sec
10 joules/sec
Physics
1 answer:
Leya [2.2K]3 years ago
7 0

Answer:

20 joules/sec

Explanation:

100/5=20

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a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu
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Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

Q_1=100\times x

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Q_2=100\times 4.186\times 11=4604.6Joules

Thus total energy needed equals Q_1+Q_2=100x+4604.6

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules

Now by conservation of energy we have

Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g

6 0
3 years ago
Which of the following is an example of rolling friction?
il63 [147K]

Answer:

O bike tires on the road as you ride

Explanation:

is the rolling friction

7 0
3 years ago
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What is one standard kilogramun si system<br><br><br><br><br>​
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Answer:

The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.

3 0
3 years ago
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A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t
beks73 [17]
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy. 

<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>

<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>

<span>∆h = 0.51 m = 51 cm </span>

<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>

<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>

<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>

<span>The Frictional energy converted is F . d </span>

<span>F ( the frictional force ) = µ . N </span>

<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>

<span>Total energy converted </span>

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5 0
3 years ago
Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal
Dmitry_Shevchenko [17]

Answer:

a = \dfrac{4\pi^2R}{T^2}

Explanation:

The acceleration of a circular motion is given by

a = \omega^2 R

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\omega=\dfrac{2\pi}{T}

Substitute into the previous formula.

a = (\dfrac{2\pi}{T})^2 R

a = \dfrac{4\pi^2R}{T^2}

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0
3 years ago
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