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3 years ago

Please view attached images. The first one is #2 and the second one is #1.

Physics

1 answer:

MakcuM [25]3 years ago

8 0

The proton would be 2 and d a part of 1 then calculate that hope this helped

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At a construction site, the site manager notices that a crane takes 20 seconds to lift a 500kg steel beam up to a height of 15 m

posledela

The work done is equal to the change in potential energy which is:
P.E = mgh
P.E = 500 x 9.81 x 15
P.E = 73,575 J

Power = work / time
Power = 73,575 / 20
Power = 3,700 Watts

5 0

4 years ago

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam

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3 years ago

Select the best answer for the question.

Aleonysh [2.5K]

Answer:

I think the answer will be water ,sorry if ik wrong

4 0

3 years ago

A green light has a wavelength of 485 nm.<br> What is its frequency? (1 nanometer is 10-9 m)

Lynna [10]

Answer:

Explanation:

C bro

8 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

4 years ago