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3 years ago

True Or False? The tendency for an object in motion to remain in motion is called orbital speed.

Physics

2 answers:

boyakko [2]3 years ago

8 0

False...I am pretty sure

Harlamova29_29 [7]3 years ago

7 0

The correct answer to the question is False i.e the tendency of an object in motion to remain in motion is not called the orbital speed.

EXPLANATION:

Before going to answer this question, first we have to understand Newton's first laws of motion.

As per Newton's first laws of motion, every body continues to be in state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces.

Hence, as long as no unbalanced force is acting on a moving object, it will be in motion. This tendency of a moving object to be in motion is called inertia of motion of the body.

Inertia of motion is the property of the body by virtue of which a moving body always tries to be in motion.

Hence, the tendency of an object in motion to remain in motion is not called as the orbital speed.

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Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

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Jacob's family celebrates with him as he jumps with both feet together. Jacob is likely (2 points)

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3 years ago

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Models of four atoms are shown as figures A, B, C, and D. Which two atoms represent isotopes of the same element?

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3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?

Serjik [45]

The image of the object is 8cm to the left of the lens (D)

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0

2 years ago