BI

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

3 years ago

What is the chief benefit of anaerobic exercise?

ra1l [238]

Answer:

c isthe answer if not see my profile m e. e. t me I will explain u

8 0

3 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the

saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

$R=\alpha \frac{l}{s}$

Were:

R is resistance

$\alpha$ is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area ( $s=\pi r^{2}$ ).

So if the lenght and radius are doubled, the equation goes as follows:

$R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }$

So finally because the circular area is a square function, the resulting equation is half of the one before.

7 0

3 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

3 years ago

An astronaut drops a hammer from 2.0 meters above the surface of the moon. if the acceleration due to gravity on the moon is 1.6

aivan3 [116]

Answer:

1.57 s

Explanation:

Since the motion of the hammer is a uniformly accelerated motion, the distance covered by the hammer in a time t is

$S=\frac{1}{2}at^2$

Where, in this case

S = 2.0 m is the distance covered

a = 1.62 m/s^2 is the acceleration due to gravity

t is the time taken

Re-arranging the equation, we can find the time the hammer takes:

$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(2.0 m)}{1.62 m/s^2}}=1.57 s$

8 0

3 years ago

Read 2 more answers