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Tom [10]
3 years ago
7

Calculate the potential energy of a 5.2 kg object positioned 5.8 m above the ground.

Physics
2 answers:
GrogVix [38]3 years ago
4 0

Answer:

295.568J

Explanation:

use P=mgh

plug in givens

P= 5.2*9.8*5.8= 295.568J

Free_Kalibri [48]3 years ago
3 0

\bf \huge \hookrightarrow \:Given :

\bf \implies \: Weight  \:  \: of  \:  \: Object \:  = 5.2 \: kg

\bf \implies  \: Height \:  = 5.8m

<h2>_________________________</h2>

\bf \huge \hookrightarrow \:To  \:  \: Find

<h3> ☛ Potential Energy of object.</h3>

<h2>_________________________</h2>

\bf \Large \hookrightarrow \:Formula  \:  \: using

\Large  \mid   \underline {\bf {{{\color{orange}{p \:  =  \: mgh}}}}} \mid

  • P denotes Potential Energy

  • m denotes mass

  • g denotes gravitational field

  • h denotes height

<h2>_________________________</h2>

<h3>☛ Now , putting the values in formula.</h3>

  • We know that , the gravitational field of earth is 9.8.

\bf \Large \mapsto \: \: 5.2 \:  \times 9.8 \:  \times 5.8 \\  \\ \bf \Large \mapsto \: \: 50.96 \:  \times 5.8 \\  \\ \bf \Large \mapsto \: \: 295.56

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a battery of 9v is connected in series with resistors of 0.2ohm , 0.3ohm, 0.5ohm, and 5 ohm respectively. how much current would
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Answer:

1.5 A

Explanation:

Applying

V = IR'....................... Equation 1

make I the subject of the equation

I = V/R'.................. Equation 2

Where V = Voltage, I = current, R' = Total resistance.

From the question,

In a series connection,

R' = 0.2+0.3+0.5+5 = 6 ohm.

Given: V = 9V

Substitute into equation 2

I = 9/6

I = 1.5 A.

Note: Since all the resistors are connected in series, thesame current flows through them

Therefore the current flowing through the 5 ohm resistor = 1.5 A

5 0
3 years ago
Changes that occur in the urinary system with aging include all of the following, EXCEPT
Sladkaya [172]

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Explanation:

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6 0
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A baseball player throws a baseball at 42 m/s. If the other player catching the baseball is 18 m from the player, how much time
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3 0
3 years ago
Now imagine a person dragging a 50 kg box along the ground with a rope, as
ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

                                                     = 50 x 9.8

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The coefficient of friction,

                            μ = f/ η

                               = 30 / 490

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Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
4 years ago
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