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Oksana_A [137]
3 years ago
7

A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

train move away from the track gang and what was its speed after the 5 min.?​
Physics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check:     V0 = 5 m/s

                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s     (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s  the speed after 300 sec

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Answer:

Thin, aluminium and buried underground.

Explanation:

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ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

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3 years ago
How is aquatic biome classification different from terrestrial biome classification?
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A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to
blagie [28]

Answer:

the density of indium is  7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass  ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

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hence, the density of indium is  7.2 g/cm^3

We simply applied the above formula so that the correct value could come

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8 0
3 years ago
What is the amount of force required to accelerate a 20 kg object to 5 m/s˛?
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Answer:

Force = 100N

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3 0
3 years ago
A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot
Reil [10]
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
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where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
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at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
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to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
6 0
4 years ago
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