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3 years ago

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Use this scenario to answer the four following questions. A person is on a sled on top of a snow covered hill. The vertical dist

ance from the top of the hill to the bottom is 12 meters. The sled and the person riding it have a combined mass of 60 kg. To start, person is pushed. The initial velocity is 1 m/s. At the bottom of the hill, the sled and the person have a velocity of 10 m/s. Use g=9.8 m/s2 or g=10 m/s2 for the gravitational acceleration. What is the initial total mechanical energy of the person and their sled?

1 answer:

jekas [21]3 years ago

7 0

Answer:7086 J

Explanation:

Given

height of hill from ground(h)=12 m

combined mass of sled and man(m)=60 kg

initial velocity(u)=1 m/s

at bottom velocity(v)=10 m/s

Total mechanical energy of the person

E=Potential Energy+Kinetic Energy

$E=mgh+\frac{mv^2}{2}$

$E=60\times 9.8\times 12+\frac{60\times 1^2}{2}$

E=7056+30=7086 J

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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)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

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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