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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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<u>Answer</u>

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<u>Explanation</u>

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10,000 cL = 1 hL

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Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

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3 years ago

When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?

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4 years ago

Which of the following best demonstrates the effect of static friction? A. A person pushing a couch and it slowly sliding across

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3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$