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yanalaym [24]
3 years ago
7

How might the temperatures on Mercury be different if it had the same mass as Earth?

Physics
1 answer:
Reil [10]3 years ago
4 0

Answer:

well, mercury is the closest planet to the sun, making it MUCH hotter than earth. that being said, just because mercury is closest to the sun, it doesnt mean its the hottest planet.

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Which of the following statements is true? A. Whenever an object moves, work is being done. B. A larger force always means more
melamori03 [73]
The Correct answer would be D. Work equals force times distance.
8 0
3 years ago
Which of the following are capital cities in the Caribbean? Lima La Habana San Juan San Jos Guinea Ecuatorial
mars1129 [50]

Caribbean:
Habana ... Cuba
San Juan ... Puerto Rico
San Jose ... Costa Rica

Other:
Lima ... capital of Peru in South America
Equatorial Guinea ... country in Africa

5 0
3 years ago
What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita
sesenic [268]

Answer:

60m/s

Explanation:

initial energy = final energy

g.p.e = k.e

k.e = 0.5 × mass × velocity²

g.p.e = 990000J as per Question

990000Nm = 0.5 × 550 × V²

V² = 3600

V = 60m/s

4 0
3 years ago
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of
seraphim [82]

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

8 0
3 years ago
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