Ask question

Ask question

All categories

2 years ago

11

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

2/c^2, what is the distance, to the

2 answers:

viktelen [127]2 years ago

9 0

0.31 for E2020 users

Nataliya [291]2 years ago

7 0

Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2 m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low

You might be interested in

Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun

fenix001 [56]

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

3 0

2 years ago

Two swimmers begin a race and Swimmer A completes each length of the pool in 30 seconds, while Swimmer B completes each length i

Anni [7]

Answer:1.33 times

Explanation:

Given

A takes 30 sec to complete each length

While B takes 40 sec to complete each length

Let L be the Length of Swimming pool Length

and $V_a$ and $V_b$ be the length of A & B

Thus after 5 minutes A traveled

$x_a=\frac{L}{30}\times 5\times 60=10L$

$x_b=\frac{L}{40}\times 5\times 60=7.5L$

so A is 2.5L units ahead of B

A has traveled 1.33 times than B in 5 mins

6 0

2 years ago

A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

(2) Angular frequency of the wave is given by :

$\omega=2\pi\nu$

$\omega=2\pi\times 4921.56\ Hz$

$\omega=30923.07\ rad/s$

(3) The period of oscillation is given by T as :

$T=\dfrac{1}{\nu}$

$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

5 0

2 years ago

A bobsled is pushed with a force of 190.08 N. The sled has a mass of 28 kg. What is the acceleration of the bobsled? Report to t

Usimov [2.4K]

By definition it is known that force equals mass by acceleration. In other words F = m * a. To find the acceleration, you must clear the formula mentioned. Therefore, for a force of 190.08N and a mass of 28 Kg, we have that the acceleration is a = F / m = (190.08) / (28) = 6.79 m / s ^ 2

6 0

2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

liq [111]

Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

5 0

2 years ago