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3 years ago

Need help on question b

Physics

1 answer:

ZanzabumX [31]3 years ago

8 0

The answer is A.

A positive charge’s electric field pushes out.

Hope this helps! -Avenging

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A point charge q = +4.50 nC moves through a potential difference ΔV = Vf − Vi = +27.0 V. What is the change in the electric pote

olganol [36]

Change in electric potential energy: 121.5 nJ

Explanation:

For a charged particle moving in an electric field, the change in electric potential energy of the particle is given by

$\Delta U = q \Delta V$

where:

q is the charge of the particle

$\Delta V$ is the potential difference between the initial and final position of the particle

For the point charge in this problem, we have:

$q=+4.50 nC$ is the charge

$\Delta V=+27.0 V$ is the potential difference

Therefore, the change in electric potential energy is

$\Delta U=(+4.50)(+27.0)=121.5 nJ$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

A Ray of light falling on rough surface follows the laws of reflection but no image of the object placed before it is C explain

stepladder [879]

Answer:

The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.

4 0

3 years ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0

2 years ago

If the density of an object is 5.2 g/cm3, and volume is 3.7 cm3, what is its mass

netineya [11]

Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3

6 0

3 years ago

Read 2 more answers

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that