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zheka24 [161]
3 years ago
6

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

Physics
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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A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

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q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

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Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

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