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tankabanditka [31]
3 years ago
7

Imagine that a bowling ball needs to be lifted 1.5 meters, and its gravitational potential energy is 90 joules. How much does th

e bowling ball weigh?
Physics
1 answer:
Vikki [24]3 years ago
8 0

Answer: 6.12 kg

Explanation:

Since Mass of ball = ? (let the unknown value be Z)

Acceleration due to gravity, g= 9.8m/s^2

Height, h = 1.5 metres

Gravitational potential energy GPE = 90J

Gravitational potential energy depends on the weight of the ball, the action of gravity and height.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

90J = Z x 9.8m/s^2 x 1.5m

90 = Z x 14.7

Z = 90/14.7

Z = 6.12 kg

Thus, the bowling ball weigh 6.12 kilograms

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3 years ago
Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di
Goshia [24]

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v =  250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

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2 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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