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3 years ago

Is it possible for an object to be in motion without any external force applied? justify

1 answer:

Rudiy273 years ago

8 0

Newton’s first law is commonly stated as:
An object at rest stays at rest and an object in motion stays in motion.
However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
This, of course, is not true. In the absence of any forces, no force is required to keep an object moving. An object (such as a ball) tossed in the earth’s atmosphere slows down because of air resistance (a force). An object’s velocity will only remain constant in the absence of any forces or if the forces that act on it cancel each other out, i.e. the net force adds up to zero. This is often referred to as equilibrium. The falling ball will reach a terminal velocity (that stays constant) once the force of air resistance equals the force of gravity.

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Briefly answer the following questions. a) A clock is mounted on the wall. As you look at it, what is the direction of the angul

Lelu [443]

Answer:

Explanation:

a ) The direction of angular velocity vector of second hand will be along the line going into the plane of dial perpendicular to it.

b ) If the angular acceleration of a rigid body is zero, the angular velocity will remain constant.

c ) If another planet the same size as Earth were put into orbit around the Sun along with Earth the moment of inertia of the system will increase because the mass of the system increases. Moment of inertia depends upon mass and its distribution around the axis.

d ) Increasing the number of blades on a propeller increases the moment of inertia , because both mass and mass distribution around axis of rotation increases.

e ) It is not possible that a body has the same moment of inertia for all possible axes because a body can not remain symmetrical about all axes possible. Sphere has same moment of inertia about all axes passing through its centre.

f ) To maximize the moment of inertia of a flywheel while minimizing its weight, the shape and distribution of mass should be such that maximum mass of the body may be situated at far end of the body from axis of rotation . So flywheel must have thick outer boundaries and this should be

attached with axis with the help of thin rods .

g ) When the body is rotating at the same place , its translational kinetic energy is zero but its rotational energy can be increased

at the same place.

3 0

3 years ago

When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at spee

Charra [1.4K]

Answer:

The correct option is C

Explanation:

From the question we are told that

The initial speed of the rocket is $v_i = 100 m/s$

The speed of the rocket engine sound is $V$

The final speed of the rocket is $v_f = 200 \ m/s$

The speed of the sound at $v_f$ would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates

3 0

3 years ago

What is the starting and final energy for a battery?

EleoNora [17]

Electrical energy is the starting and final energy of a battery

8 0

3 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

3 years ago

What is the correct order of the layers' density from lowest density to highest?

Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

6 0

3 years ago

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