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ddd [48]
3 years ago
13

) In the figure below, three charges K, L and M are placed at the corners of an equilateral triangle. If K and L are fixed, in w

hich direction would M move?

Physics
1 answer:
abruzzese [7]3 years ago
4 0

Answer:

B)

Explanation:

Negative (-) charge M will not move towards negative (-) charge K because, same charges will not attract each other in the given case

Negative (-) charge at the M tends to move towards positive (+) charge L in the direction of B) because opposite charges attract each other.

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Ben is exercising his pony on a large oval track. The long is trotting at a constant speed of 10 m/s around the track. Which sta
nlexa [21]
There is 0 acceleration because the pony is moving at a constant velocity
6 0
3 years ago
TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po
Ghella [55]

Answer:

The speed is 138852.4 \frac{m}{s}

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

F=Eq

with q the charge of the proton (1.61\times10^{-19} C). Because between parallel plates electric field is almost constant, electric force is constant too:

F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

a=\frac{F}{m}

with m the mass of the proton (1.67\times10^{-27} kg):

a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}

Now with this constant acceleration we can use the kinematic equation

v^{2}=v_{0}^{2}+2ad

with v the final speed, v_i the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}

v=138852.4 \frac{m}{s}

5 0
3 years ago
7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in
viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

6 0
3 years ago
5. Describe the sequence of mechanical
butalik [34]

Answer: idk that is a tough one!

Explanation: that is a hard question IDK

5 0
3 years ago
shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r
Ainat [17]

Answer:

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0

\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

8 0
3 years ago
Read 2 more answers
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