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3 years ago

Can someone help me please

Physics

2 answers:

enyata [817]3 years ago

3 0

Answer:

I think its distance

Explanation:

when measuring how far a p.o art u can use mm

uysha [10]3 years ago

3 0

C distance bebwbsbwnwnwnsnsnannakaka

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Rashid [163]

Alnilam is the brightest star

Absolute brightness is how bright the star actually is. Apparent brightness is how bright the star looks.

Factors that influence how bright a star looks is distance. The fact that Alnilam has the same apparent brightness of the other two stars even though it is farther away, means that it is brighter.

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3 years ago

Hat processes lead to glacial erosion ?

azamat

Yes......................

8 0

3 years ago

How long will a trip take in hours of you travel 450kmat an average speed of 80 km/hr

777dan777 [17]

5.625 hours and it is 450 divided by 80
Have A Good Day

4 0

3 years ago

5. An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and

marishachu [46]

Answer:

D. 18.60

Explanation:

By the law of conservation, the momentum is neither loss nor gained but instead transfered. When they crash into each other, and stick, they combine to create a total mass of 215 kg. Since the momentum is transfered, the two objects, combined, have a total momentum of 4000 kg-m/s. We know that momentum equals mass times velocity. You then divide 4000 by 215 and get approximately 18.6 m/s

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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

3 years ago