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4 years ago

An object that has kinetic energy must what ?

Physics

1 answer:

11Alexandr11 [23.1K]4 years ago

4 0

An object that has kinetic energy must be moving.

The formula for an object's kinetic energy is

KE = (1/2) · (the object's mass) · (the object's speed)²

As you can see from the formula, if the object has no speed, then its kinetic energy is zero. That's why kinetic energy is usually called the "energy of motion", and if an object HAS kinetic energy, then that tells you right away that it must be moving.

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What kind of friction occurs as a fish swims through water?

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It must be sliding friction, because the fish is already in motion.

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3 years ago

Which action describes an experimental investigation?

EastWind [94]

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3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

If a machine increases force it must apply the force over a long distance? True or false

Eduardwww [97]

Answer: The machine must apply the force over a shorter distance. That's because a machine doesn't change the amount of work and work equals force times distance. Therefore, if force increases, distance must decrease

FALSE

HOPE THIS HELPS

4 0

3 years ago

calculate the average acceleration of a car with a starting velocity of 0 m/s that accelerates to 9 m/ s in 3 second

ladessa [460]

Answer:

a = 3 m/s²

Explanation:

a = Δv/t

a = (vf - vi)/t

a = (9 - 0)/3

a = 3m/s²

6 0

3 years ago