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AlexFokin [52]
2 years ago
6

Which graph shows the correct relationship between kinetic energy and speed? A. Graph representing the relationship between spee

d on the x-axis and kinetic energy on the y-axis. A straight line starts at the y-axis B. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A straight line starts from the origin C. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A curve line runs parallel to the x-axis and y-axis D. Graph representing a relationship between speed on the x-axis and kinetic energy on the y-axis. A semi-curve line starts above the origin on the y-axis and curves upwards as it moves forward in speed
Physics
1 answer:
Liono4ka [1.6K]2 years ago
3 0

The proper connection between kinetic energy and speed is depicted in Graph D.

<h3></h3><h3>What is kinetic energy?</h3>

The kinetic energy (KE) is defined as one-half of the mass times multiplied by the square of velocity.

\rm KE = \frac{1}{2}mv^2

where,

KE is the kinetic energy

m is the mass of each molecule

(V) is the  velocity

As the square of the velocity is exactly proportional to kinetic energy. Consequently, the relationship between velocity and kinetic energy must be parabolic.

The proper connection between kinetic energy and speed is depicted in Graph D. Graph showing the link between kinetic energy on the y-axis and speed on the x-axis.

On the y-axis, a semi-curve line begins above the origin and ascends as it accelerates.

To learn more about the kinetic energy refer to the link;

brainly.com/question/12669551

#SPJ1

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A spring, having an unstretched length of 2ft, has one end attached to the 10lb ball. determine the angle ? of the spring if the
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4 years ago
A shell is fired from the ground with an initial speed of 1.74 ✕ 103 m/s at an initial angle of 58° to the horizontal.
ValentinkaMS [17]

Answer:

a) horizontal range s=277671.77 m

b) time the shell is in motion 301.143 s

Explanation:

Is a parabolic movement so the velocity have two components:

v = 1.74 x 10^{3}  ( \frac{m}{s} )

\alpha = 58°

v_{y} =v*Sen (\alpha )\\v_{x} =v*Cos (\alpha )

v_{y} =1740*Sen (58 )\\v_{x} =1740*Cos (58)

v_{y} = 1475.603 \frac{m}{s}\\v_{x} = 922.059 \frac{m}{s}\\

t= \frac{2*v_{y} }{g}

t= \frac{2*1475.603 }{9.8}

t= 301.143 s

v= \frac{s}{t} \\s= v_{x}*t

s= 922.059 \frac{m}{s} * 301.143 s

s = 277671.77 m

4 0
4 years ago
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are th
masha68 [24]

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

7 0
3 years ago
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