A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is r

1) The electric potential energy can be defined as the product of the electric potential and the associated charge:

where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is , and the potential energy is , thus we may rearrange the earlier formula to find the electric potential at the tip:


2) Using this same formula, if the charge changes to , the resulting electric potential will be:

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

3. the field is oriented away from the axon

Explanation:

The calculation for the electric field is illustrated below:

E = k*|q|/r²

Where:

E = electric field; k = 8.98755*10⁹ N*m²/C²; r = distance separating the field being measured from the point charge = 0.05 m; q = point charge

For a length of 0.100 m of the axon, the value of q is calculated as:

q = (5.6*10¹¹)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Therefore:

q = (5.6*10¹¹)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Consequently:

E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

The work done can be calculated using the equation:

Work = Force x Distance = Change in kinetic energy

The kinetic energy is derived using the following formula: KE = (1/2)*m*v^2
Thus, the change in kinetic energy is calculated as (1/2)*m*(Vf)^2 - (1/2)*m*(Vo)^2

Where:

Vf represents the final speed = 90 kph = 25 m/s
Vo denotes the initial speed = 72 kph = 20 m/s

By substituting in the given values:

Work = (1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which can also be represented as 2.8 x 10^5 Joules.

The correct choice among the options is A.

Answer:

0.130

Explanation:

The coefficients of static friction recorded for each trial are listed as follows:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

Adding these coefficients together results in: 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

Consequently;

the mean coefficient of static friction =

                                              =

                                              = 0.12983

The mean coefficient of static friction is 0.130

Complete Question:

Picture yourself on an aluminum ladder on the ground, attempting to fix an electrical connection with a metal screwdriver featuring a metallic handle. Since you are sweating profusely, your body has a resistance of 1.60 kΩ.

(a) If you accidentally contact the "hot" wire from the 120 V power line, what current will flow through your body?

(b) What is the amount of electrical power transferred to your body?

Answer:

(a) 0.075A

(b) 9W

Explanation:

According to Ohm's law, the voltage (V) applied to or passing through a body corresponds to the current (I) via the relationship:

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R denotes the resistance of the body

(a) As mentioned;

Due to wet conditions, the body will conduct electricity, and possesses the following values;

V = supplied voltage = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substituting these values into equation(i):

120 = I x 1600

To find I;

I =

I = 0.075A

Thus the current passing through your body is 0.075A

(b) Electrical power (P), which is expressed in Watts (W), delivered to the body is the product of current (I) and voltage (V) received. Thus:

P = I x V           ---------------------(ii)

Where;

I equates to 0.075A   [as derived above]

V is 120V     [as outlined in the question]

Plugging these values into equation (ii):

P = 0.075 x 120

P = 9W

Hence, the electrical power received by your body is 9W