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Sliva [168]
3 years ago
9

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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The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount
igor_vitrenko [27]
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7 0
3 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
A projectile is shot on level ground with a
erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

5 0
2 years ago
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