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3 years ago

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A charge of 3 nC is uniformly distributed around a ring of radius 10 cm that has its center at the origin and its axis along the

x axis. A point charge of 1 nC is located at x = 2.5 m. Find the work required to move the point charge to the origin. Hint: The electric field on the axis was calculated in class (see Example 3 in the the slides for Lecture 16).

1 answer:

Step2247 [10]3 years ago

8 0

Answer:

$W=2.592*10^{-7} J$

Explanation:

GIVEN DATA:

Charge 3 nC

Radius of ring is 10 cm

A point charge 1 nC

Distance of point charge is 2.5 m

We know that voltage is calculated as

$V(0) =\frac{KQ}{r}$

$=\frac{(9*10^{-9})(3*10^{-9})}{0.1}$

V = 270 V

At x = 2.5m

$r = \sqrt{(0.1^2+2.5^2)}$

r = 1.581 m

$V(2) = \frac{(9*10^9)(3*10^{-9})}{1.581}$

V(2) = 10.801 V

Work done is calculated as

W= q(dV)

$W = (1*10^{-9})(270 - 10.80)$

$W=2.592*10^{-7} J$

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A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

What is one watt ? Write the relation of watt with kilowatt , megawatt , and horsepower .

alina1380 [7]

Answer:

See the explanation below

Explanation:

The watt (the power) is equal to the relationship between the work and the time in which that work is performed.

$P = W/t$

where:

W = work [J] (units of Joules)

t = time [s].

Now 1000 [W] are equal to 1 [kW]

And 1000000 [W] are equal to 1 [MW]

The horsepower is the unit of power in the imperial system of units.

And 745.7 [W] are equal to 1 [Hp]

3 0

2 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

2 years ago

How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: T

pychu [463]

Answer:

33.5 kJ

Explanation:

here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.

we have ice 100g =0.1 kg

Appplying Q=mL

Q= $0.1 kg * 335 000 J kg^{-1}$

Q = 33 500 J

Q = 33.5 kJ

3 0

3 years ago

Select the correct answer.

kherson [118]

Answer:

B chemical energy

Explanation:

4 0

2 years ago