Explain the difference between the justice system laws and scientific laws

If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,

jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J

5 0

3 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m

kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

$m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}$

Put the value into the formula

$f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}$

$f_{e}=3.68\ cm$

Hence, The focal length of eyepiece is 3.68 cm.

3 0

3 years ago

An athlete runs a 100 m race in 10 seconds against a friction force of 100N. How do I work out his power output?

Rashid [163]

They seem to cancel each other out which is odd

6 0

2 years ago

Sitting in a chairlift, Rebecca has a gravitational potential energy of 5,997.6

stira [4]

Answer:

B) 12 m

Explanation:

Gravitational potential energy is:

PE = mgh

Given PE = 5997.6 J, and m = 51 kg:

5997.6 J = (51 kg) (9.8 m/s²) h

h = 12 m

8 0

3 years ago