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stellarik [79]
3 years ago
15

A car drives 400 meters in 5 seconds. What is the cars speed in m/s?

Physics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

80/1

Explanation:

to figure out the answer of how many miles/meters per second divide the meters/miles by the second

so 400÷5 would be 80

80 meters per second

hope this helps :))

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A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less tha
lakkis [162]

Answer:

p = -q  

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

            1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

          1 / p = - 1 / q

           p = -q

Therefore the distance is equal to the current distance, so the distance does not change

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3 years ago
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Why is outer space silent
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The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45
zheka24 [161]

Answer:

v = 12 m/s

Long, boring, and convoluted explanation:

First, let's lay out our information.

- <em>max height = 3.7 m</em>

- <em>0 = 45°</em>

<em>- gravitational acceleration constant = 9.8 </em>\frac{m}{s^2}<em />

<em />

Since the puma leaves the ground at a <em> 45 °</em> angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is  <em>3.7m</em>  because the puma can reach up to that height. Gravity is always given to be <em>9.8 </em>\frac{m}{s^2}<em />

<em />

Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity,  <em>vsin ( θ )</em> , and not the horizontal component, <em>vcos ( θ )</em> .

Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: <em>vsin ( θ )  = </em> \sqrt{2hg}

Where <em> vsin ( θ )</em>  is the vertical component of the initial velocity and <em>h</em>  and <em>g</em> are max height and gravitational acceleration constant respectively.

Plugin, rearrange and solve

v sin ( θ )  =  \sqrt{2hg}

v sin ( 45 ∘ )  =   √ 2  ×  3.7  ×  9.8

v ( 0.71 )  =  \sqrt{72.52}

v ( 0.71 )  =  8.52

v  =  8.52 /0.71

v =  12 m s

<em />

<em />

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Suppose you are investigating how the day ends hamsters affect how much they run on the hamster wheel. In order to control the e
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The answer is C because you need non running hamsters to compare running hamsters with.
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