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disa [49]
2 years ago
6

Hey guys, I don't know what this is. Any help?

Physics
1 answer:
Marizza181 [45]2 years ago
4 0
Real and reduced is the answer
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A runner makes one lap around a 200m track of 25s. What were the runners average speed and average velocity?
kobusy [5.1K]
Here is the answer. In order for us to get the speed, we are just going to use the formula speed is equal to distance over time. Given that the distance s 200m and the time is 25s, this would give us the answer of 8m/sec. Hope this answers your question.
5 0
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while racing on a flat track, a car rounds a curve of 28m radius and instantaneously experiences a centripetal acceleration of 1
liubo4ka [24]
When a body strictly moves on a curve, it's velocity at a point is tangential to the curve at that point.

Centripetal acceleration is the acceleration that a body experiences by the virtue of change in it's tangential velocity. It is directed towards the centre and mathematically is v^2/R where v is the speed at the instant.

So, 18 = v^2/R
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3 years ago
Waves begin to "feel bottom" when the depth of water is
LekaFEV [45]
I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.
8 0
3 years ago
Read 2 more answers
PLS HELP 30 POINTS IF U HELP
USPshnik [31]

Answer:

its D so yeah.........

Explanation:

tell me if it's wrong

8 0
2 years ago
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A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
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