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Simora [160]
2 years ago
5

If we assume a typical efficiency for the body, as quoted in the video, how much energy input would it take to climb a ladder, i

ncreasing your body’s potential energy by 5,000 j?.
Physics
1 answer:
torisob [31]2 years ago
8 0

The energy input would it take to climb a ladder will be 20000 J.

<h3>What is the mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of the given mechanical machine.

Mechanical advantage is a measure of how much a machine multiplies the input force.

The complete question is;

"if we assume a typical efficiency of 25% for the human body how much energy input would it take to climb a ladder increasing body potential energy by 5000J"

The formula for the efficiency is;

\rm E = \frac{P}{\eta} \\\\ E = \frac{5000}{0.25} \\\\ E = 20000  \ J

Hence, the energy input would it take to climb a ladder will be 20000 J.

To learn more about the mechanical advantage, refer to the link;

brainly.com/question/7638820

#SPJ1

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Two sacks contain the same number of identical apples and are separated by a distance r. The two
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Answer:

The appropriate response will be "F_2=\frac{3}{4}F". A further explanation is given below.

Explanation:

According to the question,

⇒ F_1=\frac{G(m_1 m_2)}{r^2} ....(equation 1)

and,

⇒ F_2=\frac{G(m_1 m_2)}{r^2} ....(equation 2)

Now,

On dividing both the equations, we get

⇒ \frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}

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⇒ F_2=\frac{3}{4}F

4 0
3 years ago
What will be the answer??
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5-a).  Acceleration is a vector defined as the rate of change of velocity.
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Its direction is the direction in which velocity is increasing. 

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When we look at any time, from zero to almost 50 minutes, the
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4 years ago
Having the correct posture and the degree or severity of braking, acceleration and steering inputs have a direct result on the a
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3 years ago
A parallel plate capacitor has a capacitance of 6.78 μF and it is connected to a 12V battery. What is the charge on the capacito
AfilCa [17]

Answer:

8.136×10⁻⁵ J

Explanation:

Applying,

Q = Cv................ Equation 1

Where Q = Charge on the capacitor, v = voltage of the battery, C = capacitance of the capacitor.

From the question,

Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V

Substitute these values into equation 1

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