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Natali [406]
3 years ago
12

Help please:A boy weighing 30 kg rides a scooter. The total kinetic energy of the boy and the scooter is 437.6 J. Determine the

speed at which the boy moves if the mass of the scooter is 5kg?​
Physics
2 answers:
Ksenya-84 [330]3 years ago
5 0

Answer:

the total mass is 35 kg

k.E = 1/2 mv2

43.76 =1/2 v2

v2=2×43.76

Explanation:

v=radicls 87.52 which is equal to 9.5 m/s

larisa86 [58]3 years ago
3 0

Answer:

5 m/s

Explanation:

Mass of the boy + scooter system =  35 kg

Kinetic Energy =  437.6 J

437.6 = \frac{1}{2} 35 {v}^2

25.0057143 = v^{2}

v = 5 m/s

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Answer:

Explanation:

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The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α    

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(T2 - T1)R = 0.5 x m3 x R² x a / R

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from (1), (2) and (3)

a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g

a = \frac{4.5-3}{3+4.5+0.75}\times 9.8

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

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Any four difference between velocity and acceleration​
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Which type of communications equipment functions as a radio receiver and searches across several frequencies?
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If a block of wood has a density of 0.6g/cm3 and a mass of 120g what is the volume
Luda [366]

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3 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

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v = Velocity

k = Spring constant

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

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(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

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Replacing,

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v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

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