All categories

3 years ago

What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.

2 answers:

soldier1979 [14.2K]3 years ago

7 0

The kinetic energy of an object is
$\frac{1}{2} mv^2$
where m is the mass of the object
and v is the velocity.

If you roll a snowball downhill, gravity accelerates the object and v increases. This causes the kinetic energy of the object to increase as well.

If you roll a snowball on level ground or uphill, friction will cause the velocity to decrease. Thus kinetic energy will decrease as well.

LiRa [457]3 years ago

5 0

If the ground is level, then the snowball can never have
any more kinetic energy than it hand when it left your hand.

If more mass sticks to it as it makes its way across the lawn,
then it must slow down, so that its

KE = (1/2) (present mass) (present speed)²

never exceeds the KE you gave it when you tossed it.

And we're not even talking yet about all the energy it loses
by scraping through the snow and mashing down the blades
of grass in its path.

You might be interested in

When you throw a ball upward, its kinetic energy ____ (DECREASES,INCREASES,REMAINS THE SAME) and its potential energy _____ (DEC

Anettt [7]

When the ball goes upward, the velocity is decreasing. Therefore, the kinetic energy is decreased. The potential energy when the ball's height increases also increases. When the ball reaches maximum, the velocity becomes zero and the kinetic energy is minimum.

6 0

3 years ago

Read 2 more answers

A fence 8 ft high (w) runs parallel to a tall building and is 24 ft (d) from it. Find the length (L) of the shortest ladder t

AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

$Hypotenuse^2 = opposite^2 + adjacent^2$

The hypotenuse is the length of the ladder.

$Hypotenuse = \sqrt{opposite^2 + adjacent^2}$

$L^2 = BC^2 + (24 + AE)^2$ ........1

Triangle ABC is similar to triangle AEF, hence:

$\frac{BC}{8} = \frac{AE + 24}{AE}$

BC = $\frac{8(AE + 24)}{AE}$ .................2

Substitute 2 into 1

$L^2$ = $(\frac{8(AE + 24)}{AE})^2$ + $(24 + AE)^2$

Let AE = x

$L^2$ = $(\frac{8x + 192}{x})^2 + (24 + x)^2$

= $(8 + \frac{192}{x})^2 + (24 + x)^2$

Minimize L with respect to x.

2 $\frac{dL}{dx}$ = $2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)$

5 0

2 years ago

Is the energy of a wave is calculated by squaring the frequency of the wave?

Misha Larkins [42]

Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!

5 0

2 years ago

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

kirill [66]

Answer:

U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

dU = - F. dr

the esxresion for strength is

F = B / r³

let's replace

∫ dU = - ∫ B / r³ dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

U = B / 2r²

we substitute the value of B = 2

U = 1 / r²

5 0

2 years ago

17. (16 pts) You set up a two-slit experiment with a laser that produces light with a

Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

. ie. for π difference it is 300nm

4 0

3 years ago