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3 years ago

15

A positive charge is moved from point a to point b along an equipotential surface. How much work is performed or required in mov

ing the charge?

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

The work done is zero

Explanation:

No work is performed or required in moving the positive charge from point A to point B

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At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy

Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using the work-energy principle

work done = change in kinetic energy

$W=\Delta K.E$

$\mu\ mg\times d=\dfrac{1}{2}mv^2$

$v^2=2\mu g d$

Put the value into the formula

$v=\sqrt{2\times0.42\times9.8\times88}$

$v=26.91\ m/s$

Hence, The the speed of the car is 26.91 m/s.

3 0

3 years ago

Read 2 more answers

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ozzi

Answer:

$r=2.4m$

Explanation:

We have to use the centripetal force equation

$Fc=\frac{mv^{2} }{r}$

we need the radious so we have to isolate "r" and we get

$r=\frac{mv^{2} }{Fc}$

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

$r=\frac{65*4.1^{2} }{455}$

$r=2.4m$

The radius of the amusement park chamber is 2.4m

4 0

3 years ago

Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted

kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

#PLATOLIVESMATTER

7 0

2 years ago

What is solar energy?

Flura [38]

I think it's D because sunlight is a solar system

3 0

3 years ago

In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

soldier1979 [14.2K]

Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ( $\lambda_{0} = 121.6nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Then, for this particular case it is gotten:

Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

Star 1:

$Blueshift: 120.4nm < 121.6nm$

Toward us

Star 2:

$Blueshift: 121.4nm < 121.6nm$

Toward us

Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

$Redshift: 122.9nm > 121.6nm$

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

<em>Case for star 1 $\lambda_{measured} = 120.4 nm$ :</em>

<em></em>

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 $\lambda_{measured} = 121.4 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

<em>Case for star 3 $\lambda_{measured} = 122.2 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

<em>Case for star 4 $\lambda_{measured} = 122.9 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

4 0

3 years ago