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KatRina [158]
2 years ago
13

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

that must be applied at it's other hand to keep it in horizontal equilibrium is.​
Physics
1 answer:
solniwko [45]2 years ago
3 0

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

           30(30) = F(45)

                    F = 900/45 = 20 N

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Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = 9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J

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2 years ago
At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops
jasenka [17]

Answer:

force = 11.33 kg-m/s^{2}

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_{1}=mU

P_{1}= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=\frac{\Delta P}{\Delta t}

F = \frac{P_{1}-P_{2}}{T}

Kgm/s^2

F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}[/tex]

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3 years ago
If it takes 15 minutes to dry your hair with a 1.200 kW hair drier, how much energy is used in drying your hair?
Elodia [21]
1.2 kW * 0.25 h = 0.300 kWh
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3 years ago
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A wave has a wavelength of 10 mm and a frequency of 14 hertz. What is its speed?
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0.14 meters per second .
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A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,
Trava [24]

Given data:

* The mass of the ball is 2 kg.

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Solution:

The weight of the ball on the planet X is,

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Substituting the known values,

\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}

Thus, the weight of the ball on the surface of planet X is 10 N.

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1 year ago
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