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viva [34]
3 years ago
7

On Earth, a scale shows that you weigh 585 N. What would the scale read on the Moon (g = 1.60 m/s2)? N

Physics
1 answer:
Serggg [28]3 years ago
8 0

Answer:

95.51 N

Explanation:

First, find the mass in kg:

Fg = 585 N

Fg = m*g

585 N = m*9.8 m/s^2

<u>m = 59.69 kg</u>

Then, to find your weight (Fg) on the moon, you use the same equation of

Fg(moon) = m*g, except this time g = 1.60 m/s^2

Fg(moon) = 59.69 kg * 1.60 m/s^2

Fg(moon) = 95.51 N

Hope this helps!! :)

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Magnitude of fourth displacement is approximately 95 metres,

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The chart shows data for an object moving at a constant acceleration. Which values best complete the chart? Time (s) Velocity (m
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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

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QE = F_x

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since string makes 30 degree angle with the vertical so we will have

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5 0
3 years ago
Please help! Will give brainliest. 
I am Lyosha [343]

<span>The correct frequency when you tune a guitar is when you hear the right tune in your own hearing and standard. The measure frequency of a guitar string is when you measure the tune of the string correctly. This is not the same because manual tuning is affected by many factors.</span>

9 0
3 years ago
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
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