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3 years ago

A train travels 225 kilometers in 2.5 hours . what's the trains average speed

2 answers:

8 0

Answer=90 kilometers per hour

225 km 2.5hours

225km/2.5hours=90 kilometers per hour

The average speed is 90 kilometers per hour

Maslowich3 years ago

7 0

Let us first concentrate and write down all the information's already given in the question. Based on the information's the answer to the question can be easily deduced.
Distance traveled by the train = 225 km
Time taken to travel the distance of 225 km = 2.5 hours
Then
Average speed = Distance traveled/Time taken
= 225/2.5 km/hour
= 90 km/hour
So the average speed of the train is 90 kilometers per hour. The procedure for doing this kind of problem is described and it is the simplest way of solving such problems. I hope you have understood the process and got the answer you were looking for.

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What is the surface pressure of Venus?

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answers 93 bar

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3 years ago

On a frictionless track,cart 1 is moving with a constant,rightward(+) velocity of 1.0m/s.Cart 2 is also moving rightward with co

mamaluj [8]

Answer:

The final velocity of cart 1 is 3m/s

Explanation:

From principle of conservation of linear momentum, which states that sum of the momentum before collision is equal to the sum of the momentum after collision.

Momentum, P is given as mass x velocity.

ΔP = Δmv = m₁u₁ +m₂u₂ = m₁v₁ + m₂v₂

Assumptions:

If the two carts are moving on frictionless track, then limiting frictional forces due to their weights are negligible.

After the elastic collision, the two carts will move separately with different velocity

u₁ + u₂ = v₁ + v₂;

where;

u₁ and u₂ are the initial velocity for cart 1 and cart 2 respectively

v₁ and v₂ are the final velocity for cart 1 and cart 2 respectively

1 m/s + 5 m/s = v₁ + 3m/s

6 m/s = v₁ + 3m/s

v₁ = 6 m/s - 3m/s = 3m/s

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3 years ago

On what factors, current sensitivity voltage sensitivity of a galvanometer depend?

Eddi Din [679]

Answer:

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3 years ago

(a) A 70-kg person at rest has an oxygen consumption rate Qhum = 14.5 liter/h, 2% of which is supplied by diffusion through the

nevsk [136]

Answer:

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Explanation:

(a)

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area of human skin, $A_h=1.7\ m^2$

diffusion rate through skin of humans, $d=2\%\ of\ Q_h$

∴ $d=\frac{2}{100} \times 14.5$

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Flux of diffusion rate,

$fd=\frac{d}{A}$

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$fd=1.7058\times 10^{-5}\ L.hr^{-1}.cm(b)Surface area for a spherical animal:[tex]A=4.\pi.r^2$

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$fd=\frac{14.5}{A}$

$1.7058\times 10^{-5}=\frac{14.5}{4.\pi.r^2}$

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Explanation:

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