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2 years ago

How much force is required to accelerate an 8.6kg wagon by 15 m/s ?

Physics

1 answer:

OverLord2011 [107]2 years ago

7 0

The amount of force required to accelerate the given mass of the wagon is 129 Newtons.

<h3>What is force?</h3>

A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.

From Newton's Second Law, force is expressed as;

F = m × a

Where is mass of object and a is the acceleration.

Given the data in the question;

Mass of the rock m = 8.6kg

Acceleration a = 15m/s²

Force F = ?

F = 8.6kg × 15m/s²

F = 129kgm/s²

F = 129N

Therefore the amount of force required to accelerate the given mass of the wagon is 129 Newtons.

Learn more about force here: brainly.com/question/27196358

#SPJ1

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3 years ago

Read 2 more answers

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

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3 0

2 years ago

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within

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Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0

2 years ago

An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionles

Arada [10]

Answer:

See below

Explanation:

a) Spring force at release = k * d = 132 N/m * .120 m = 15.84 N

b) F = ma

15.84 = (.660 kg)(a) a = 24 m/s^2

c) Toward the left ....the object is accelerated to the left

7 0

1 year ago