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Scilla [17]
2 years ago
9

An unknown fluid flows at a speed of 31 m/s. Suppose the fluid has a mass of 47 kg runs at this speed. What is the fluid’s kinet

ic energy?
Physics
1 answer:
Leya [2.2K]2 years ago
7 0

Answer:

22583.5J

Explanation:

KE=1/2 mv^2

=1/2*47Kg*(31m/s^2)

=23.5Kg * 961m/s^2

=22583.5J

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WILL MARK AS BRAINLIEST IF CORRECT ANSWERS PLZ HELP 15 PTS!!!!!!
nlexa [21]

Question 9 is 100 N.

Question 10 is No, the bike's motion is not changing so it could be at rest or moving at a constant velocity.

Question 11 is Be doubled.

Question 12 is Ella is correct.

Hope i helped.

6 0
3 years ago
Read 2 more answers
A tennis ball is dropped from 1.13 m above the
Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0
3 years ago
A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1
alexgriva [62]

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) =  2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

5 0
3 years ago
the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



6 0
3 years ago
2) A constant net force acts on an object. Describe the motion of the object.?
77julia77 [94]
Constant acceleration
7 0
3 years ago
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