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2 years ago

How much potential energy does a 1kg mess have 10m off the ground?

Physics

1 answer:

777dan777 [17]2 years ago

3 0

Potential Energy = mass x gravitational acceleration x height
potential Energy = 1 x 9.8 x 10 = 98 joules

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An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$

where:

$F_a = 15 N$ is the applied force, pushing forward

$F_f = 5 N$ is the frictional force, acting backward

$mg sin \theta$ is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=15^{\circ}$ is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

$a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

2 years ago

A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

nlexa [21]

Answer:

a) w = 2.57 rad / s , b) α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

Em₀ = U = m g h

Final. Touching the ground

$Em_{f}$ = K = ½ I w²

How energy is conserved

Em₀ = $Em_{f}$

mg h = ½ I w2

The moment of specific object inertia

I = m L²

We replace

m g h = ½ (mL²) w²

w² = 2g h / L²

The height of the object is the length of the bar

h = L

w = √ 2g / L

w = √ (2 9.8 / 2.97)

w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

τ = I α

W L = I α

mg L = (m L²) α

α = g / L

α = 9.8 / 2.97

α = 3.3 rad / s²

3 0

2 years ago

Two charges, -20 C and +4.7 C, are fixed in place and separated by 3.0 m. a. At what spot along a line through the charges is th

Pani-rosa [81]

The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.

Explanation:

Let at A both net electric field is zero then

At A ,E1=E2

E1=k*Iq1I / (d+x)^2

E2=k*Iq2I /x^2

Equating both