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Ratling [72]
4 years ago
7

13. A set of pulleys lifts an 800 N crate 4 meters in 7 seconds. What power was used?

Physics
2 answers:
kiruha [24]4 years ago
8 0

Answer:power=457.1watts

Explanation:

Vlad [161]4 years ago
5 0

Answer:

457.14 Watt

Explanation:

The information we have is:

Force: F=800N

Distance: d=4m

Time: t=7s

the formula for the power is:

P=\frac{W}{t}

where W is work, and t si time.

Also, the formula for work is:

W=F*d

Where F is the force, and d is the distance.

Substituting the formula for work into the formula for power:

W=\frac{F*d}{t}

substituting all the values to find the power:

P=\frac{(800N)(4m)}{7s}=\frac{3200Nm}{7s}\\ P=457.14W

the power that was used is 457.14 Watt

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Olympus Mons on Mars.

Explanation:

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For reference, Mount Everest stands at 8.8K.

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A plant uses the energy from sunlight to create oxygen and glucose from carbon dioxide and water. Which statement is true about
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3 years ago
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Sobre un cuerpo de 700 g de masa que se apoya en una mesa horizontal se aplica una fuerza de 5N en la dirección del plano. Calcu
kicyunya [14]

Answer:

(a) Ff = 3.95N

(b) Ff = 5N

Explanation:

(a) To find the friction force you use the second Newton law:

F=ma

But you take into account the applied force and the opposite friction force:

F_a-F_f=ma

you do Ff the subject of the formula and replace the values of the parameters:

F_f=F_a-ma=5N-(0.700kg)(1.5m/s^2)\\\\F_f=3.95N

(b) In the case of a motion with constant velocity you have that there is no acceleration. Hence:

F_a-F_f=0\\\\F_f=F_a\\\\F_f=5N

6 0
3 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

5 0
3 years ago
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