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2 years ago

8

The Earth moving round the Sun in a circular orbit is acted upon by a force, and hence work must be done on the Earth by this fo

rce. Do you agree with this statement?

2 answers:

swat322 years ago

6 0

No the earth moves around the sun in an elliptical orbit..
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ira [324]2 years ago

6 0

False, the Earth is actually orbiting the sun in a elliptical orbitation

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Arisa [49]

Answer:

Each type of electromagnetic radiation have been matched to their descriptions as follows:

Ultraviolet light A) Given off by very hot objects, such as the sun

Gamma rays B) Given off by radioactive substances

Microwaves C) Used in radar and to heat food

X‒rays D) Used in the medical field to see bone structure

Explanation:

PLS GIVE BRAINLEST

6 0

2 years ago

A ball is rolled of a 2.14m table at 8m/s. Find the time of flight for the ball and the horizontal range

Mama L [17]

Answer:

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²

Δx = 5.29 m

Round as needed.

3 0

2 years ago

A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a

prohojiy [21]

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

6 0

2 years ago

Read 2 more answers

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

A jetliner flies at a constant speed covering 467 miles in 3.3 hours. What is the speed of the plane in miles per hour?

stellarik [79]

Answer:

141.152 miles per hour is the speed of the plane in miles per hour

Explanation:

Speed of plane = Total distance travelled/total time taken -

v = D/t

Substituting the given values in the above equation, we get

v = 467/3.3 miles /hour

v = 141.152 miles per hour

141.152 miles per hour is the speed of the plane in miles per hour

8 0

2 years ago