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svlad2 [7]
2 years ago
12

If a person weighs 1000 N on the surface of the Earth, how much will they weigh on the surface of the Moon? What “g” value shoul

d we use to calculate the weight using Fg = mg?
Physics
1 answer:
Dafna1 [17]2 years ago
5 0

Answer:

Hence the weight of the person on the moon is 162.4, and the value of g used is 1.624 m/s²

Explanation:

from the question,

W = mg........................ Equation 1

Where W = weight of the man on Earth, m = mass of the man, g = acceleartion due to gravity of the man

make m the subject of the equation

m = W/g.............. Equation 2

Given: W = 1000 N,

Constant: g = 10 m/s²

Therefore,

m = 1000/10

m = 100 kg

Weight on the moon

W' = mg'

W' = 100(1.624)

W' = 162.4 N.

Hence the weight of tthe person on the moon is 162.4, and the value of g used is 1.624 m/s²

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A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and
BARSIC [14]

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

V_{0y} = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + V_{0y}²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

V_y =  V₀ × sin(θ₀) - g×t

When V_y = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

V_y =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.

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A sprinter starts from rest and accelerates to 10. m/sec in a time interval of 1.0 seconds, Calculate the runner's acceleration
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The runner's acceleration during this time interval is 10 m/s^2

<u>Given the following data:</u>

  • Initial velocity, U = 0 m/s (since the sprinter is starting from rest).
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  • Time, t = 1.0 seconds.

To calculate the runner's acceleration during this time interval, we would use the first equation of motion;

Mathematically, the first equation of motion is calculated by using the formula;

V = U + at

<u>Where:</u>

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Substituting the given parameters into the formula, we have;

10 = 0 + a(1)\\\\10 = a

Therefore, the runner's acceleration during this time interval is 10 m/s^2

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8 0
2 years ago
A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the cente
Vlad [161]

Answer:

The length of the side of the hole in the second cardboard sheet is L_2 = 0.01m

Explanation:

From the question we are told that

     The distance of the point source  from the screen is  d = 1.0 m

      The length of a side of the  first square hole is  L_1 = 0.020 \ m

      The distance of the cardboard from the point source is D_1 = 0.50\  m

   The distance of the second cardboard from the point source is D_2 = 0.25 \ m

   

Let take the  \alpha_{max } as  the angle at which the light is passing through the edges of the cardboards square hole

     Since the bright square casted on the screen by both  square holes on the   individual cardboards are then it means that

              \alpha_{max} __{1}} = \alpha_{max} __{2}}

This implies that

             tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})      

Looking at this from the SOHCAHTOA concept

               tan (\alpha_{max} __{1}}) =  \frac{opposite}{Adjacent}

     Here opposite is  the length of the side of the  first cardboard square hole

     and    

      Adjacent is  the  distance of the from the  first cardboard square hole to the point source

And for  

            tan (\alpha_{max} __{2}}) =  \frac{opposite}{Adjacent}

    Here opposite is  the length of the side of the  second  cardboards square hole (let denote it with L_2)

and

Adjacent is the distance of the from the  second  cardboards square hole to the point source

         So

                 tan (\alpha_{max} __{1}}) =  \frac{0.020}{0.50}

         And  

                tan (\alpha_{max} __{2}}) =  \frac{L_2}{0.25}

Substituting this into the above equation

                 \frac{0.020}{0.50}  =   \frac{L_2}{0.25}

Making L_2 the subject

                   L_2 = \frac{0.25 *0.020}{0.50}

                 L_2 = 0.01m

Since it is a square hole the sides are the same hence

The length of the side of the hole in the second cardboard sheet is L_2 = 0.01m

6 0
3 years ago
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