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Tomtit [17]
2 years ago
13

What time will the lunar eclipse happen central time?.

Physics
1 answer:
alexandr402 [8]2 years ago
7 0
It was about 9:30 p.m. sorry if the answer is wrong
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Multi-Select Question
dalvyx [7]

Answer:

Correct Options :

<u>Acceleration on Earth: 1.6 m/s</u>

<u>Mass on the Moon: 100 </u><u>kg</u>

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4 years ago
Equation of uniformly accelerated motion​
Kruka [31]

Answer:

x₂ = x₁ + v₁t + at²/2

Explanation:

Right out of the textbook.

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4 years ago
A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp
dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A)   What is its speed after falling for 2.00s?

  from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

  from the equation of motion s=ut+0.5at^{2} we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}

v = 15.4 m/s

7 0
3 years ago
How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at
amid [387]
Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
5 0
3 years ago
Two point charges are placed on the x axis as follows: charge q1=+3.75nc is located at x=0.205m and charge q2=−5.60 nc is at x=+
Mademuasel [1]
I might have did mistake with calculations but this is how you should do.

6 0
3 years ago
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