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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.

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When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the

Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

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We know that resistance is given by $R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega$

Length of the copper wire l = 1 m

Resisitivity of the copper wire $\rho =1.72\times 10^{-8}\Omega m$

We know that resistance $R=\frac{\rho l}{A}$

$0.00177=\frac{1.72\times 10^{-8}\times 1}{A}$

$A=969.4545\times 10^{-8}m^2$

As area $A=\pi r^2$

$3.14\times r^2=969.4545\times 10^{-8}$

$r=17.57\times 10^{-4}m$

So diameter $d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m$ = 351.42 mm

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3 years ago

Why is it better to wear a white shirt on a hot summer day instead of a black shirt?

Vadim26 [7]

Answer:

Explanation:

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3 years ago

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3 years ago

A 6.00 g lead bullet traveling at420 m/s is stopped by a large

Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

$K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}$

with m the mass and v the velocity.

$K=529.2 J$

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

$Q=\frac{K}{2}= 264.6\,J$