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Gnoma [55]
3 years ago
11

A particle of charge 3.53×10 ​−8 ​​ C experiences a force of magnitude 6.03×10 ​−6 ​​ N when it is placed in a particular point

in an electric field. What is the magnitude of the electric field at that point?
Physics
1 answer:
Cloud [144]3 years ago
6 0
<h2>Electric field at the location of the charge is 169.97 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have           

       E=\frac{F}{q}\\\\E=\frac{6\times 10^{-6}}{3.53\times 10^{-8}}\\\\E=169.97N/C

Electric field at the location of the charge is 169.97 N/C

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A sailcraft is stalled on a windless day. A fan is attached to the craft and blows air into the sail which bounces backward upon
mylen [45]

Answer:

Impulse = change in momentum w bounce

There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.

7 0
2 years ago
A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0
3 years ago
Read 2 more answers
A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

3 0
3 years ago
Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time  = 4s

Momentum  = 1000kgm/s

Unknown:

Force  = ?

Solution:

To solve this problem, we use Newton's second law of motion;

      Ft  = Momentum

F is the force

t is the time

So;

          F x 4 = 1000kgm/s

          F  = 250N

6 0
3 years ago
Two wheels having the same radius and mass rotate at the same angular velocity ((Figure 1) ). One wheel is made with spokes so n
liubo4ka [24]

Answer:

E. The wheel with spokes has about twice the KE.

See explanation in: https://quizlet.com/100717504/physics-8-mc-flash-cards/

6 0
3 years ago
Read 2 more answers
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