PHYSICS COLLEGE

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.89 Fcapacitor charged to 60.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 431, with one possessing a dielectric constant of 947.Required:
a. Find the electric potential energy of the original capacitor when it is charged. (in Joules)
b. Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

Answers

Answer 1

Answer:

$U = 3.505 *10^9 \ J$

$U_1 = 7.696 *10^9 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 1.89 \ F$

The voltage is $V = 60.9 \ k V = 60.9 *10^(3) \ V$

The first dielectric constant is $\epsilon_1 = 431$

The second dielectric constant is $\epsilon_2 = 947$

Generally the electric potential energy is mathematically represented as

$U = (1)/(2) * C * V^2$

=> $U = (1)/(2) * 1.89 * (60.9 *10^(3))^2$

=> $U = 3.505 *10^9 \ J$

Generally the capacitance when the capacitor's filling was changed is

$C_n = 1.89 * (947)/(431)$

=> $C_n = 4.15$

Generally the electric potential energy when the capacitor's filling was changed is

$U_1 = (1)/(2) * C_1 * V^2$

=> $U_1 = (1)/(2) * 4.15 * (60.9 *10^(3))^2$

=> $U_1 = 7.696 *10^9 \ J$

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What is the volume of a cone with a radius of 3 feet and a height of 6 feet use 3.14 for pie round your answer to the nearest hundredth

Answers

Answer:

56.52 feet³ to the nearest hundredth

Explanation:

the volume of a cone is given as

V =

$(1)/(3) \pi r^(2) h$

the radius is 3 feet

height is 6 feet

substituting this given values in the formular

we have that, V = $(1)/(3)$ x 3.14 x $3^(2)$ x 6

dividing , we have the volume (V)

V= 3.14 x 3 x 6

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The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer:

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K = 50N/m

New length = 32cm

Unknown

Force applied = ?

Solution:

The force applied on a spring can be derived using the expression below;

Force = KE

k is the spring constant

E is the extension

extension = new length - original length

extension = 32cm - 22cm = 10cm

convert the extension from cm to m;

100cm = 1m;

10cm will give 0.1m

So;

Force = 50N/m x 0.1m = 5N

Final answer:

To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.

Explanation:

The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).

The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.

Learn more about Hooke's Law here:

brainly.com/question/32317230

#SPJ11

Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. What is the velocity in problems per hour? a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour

Answers

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity in problems per hour will be as

velocity in problems per hour = $(complete\ problem)/(time\ taken)$ ..................1

put here value we will get

velocity in problems per hour = $(6)/(1.5)$

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Answer:

option (b) is correct

Explanation:

time t = 1 and half hour = 1.5 hour

Number of problems, n = 6

So, the velocity in problems

v = n / t

v = 6 / 1.5

v = 4

So, the rate is 4 problems per hour.

How can scientific method solve real world problems examples

Answers

The scientific method is nothing more than a process for discovering answers. While the name refers to “science,” this method of problem solving can be used for any type of problem

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm What is the distance between the two second-order minima?

Answers

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=(y_1)/(2)\n\Rightarrow \beta_1=(1.4)/(2)\n\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=(m_1\lambda D)/(d)\n\Rightarrow d=(m_1\lambda D)/(\beta_1)$

For second order

$\beta_2=(m_2\lambda D)/(d)\n\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\n\Rightarrow \beta_2=(m_2)/(m_1)\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\n\Rightarrow y_2=2(m_2)/(m_1)\beta_1\n\Rightarrow y_2=2(2)/(1)* 0.7\n\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.