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postnew [5]
3 years ago
5

What is the acceleration of a boy on a skateboard if the net force on the boy is 15 N? The total mass of the boy and the skatebo

ard is 58 kg.
Physics
2 answers:
Stels [109]3 years ago
7 0
                                 Force = (mass) x (acceleration)

If the full 15N is pointing parallel to the ground,
then

                                   15 N  =  (58 kg) x (acceleration).

Divide each side
by  58 kg:                   Acceleration = 15 N / 58 kg

                                                         = (15 kg-m/s²) / (58 kg)

                                                         = (15/58) (kg-m/kg-s²)

                                                         =  0.26 m/s² .
masya89 [10]3 years ago
6 0
Force = mass*acceleration
15 = 58 * x
x = 15 / 58
x = 0.26m/s²
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alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

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and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

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Vertically the forces acting on the object:

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so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

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now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
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and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

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V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

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