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3 years ago

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four equal charges of +3microcolumb are placed at the four corners of a square that is 40cm on a side. Find the force on any one

of the charge.

1 answer:

Reil [10]3 years ago

6 0

Answer:

i can see your question

pls fast

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Two students measure the time constant of an RC circuit. The first student charges the capacitor using a 12 V battery, then lets

luda_lava [24]

Answer:

The answer is: c.) Both students get the same time constant, since the time constant does not depend on the charge on the capacitor

Explanation:

Both students, because the time constant is not dependent on the capacitor charge. We can express the equation of the time constant as follows:

Time constant = RC

In this equation it is observed that the time constant is equal to the multiplication of the resistance (R) multiplied by the capacitance (C)

8 0

3 years ago

What is reflexive property in algerbra?

Orlov [11]

Answer: In algebra, the reflexive property of equality states that a number is always equal to itself.

hope this helps

plz mark brainliest

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3 years ago

Read 2 more answers

A 23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position o

lara [203]

Answer:

Explanation:

We use the harmonic motion position equation:

$x(t) = A\cos(\omega t+\phi)$

where A = 0.350 and for t = 0

$x(0) A = A\ cos(\phi)$

so: $\phi = 0$

and also:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s$

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

3 0

3 years ago

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

padilas [110]

Answer:

$f_{police}=1268.7 Hz$

Explanation:

We can use Doppler equation to find the frequency of the siren.

First of all we have the police car moving behind the car. Hence, the frequency detected by the car will be:

$f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})$ (1)

Now, when the police car is moving in front of the car, the frequency detected by the car will be:

$f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})$ (2)

By solving equation (1) and equation (2) for $v_{police}$ we have:

$v_{police} = 44.67 m/s$

Knowing that:

f(car1) = 1310 Hz
f(car2) = 1240 Hz
Vs = 343 m/s
V(car) = 35 m/s

Finally, we just need to put this value into the first equation to find frequency of the police car.

$f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})$

$f_{police}=1310(\frac{343-44.7}{343-35})$

$f_{police}=1268.7 Hz$

I hope it helps you!

7 0

3 years ago

A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.7 m/s2. (a) if the tension in the cord is 63

zubka84 [21]

(a)
The formula is:
∑ F = Weight + T = mass * acceleration

as the elevator and lamp are moving downward, I choose downward forces to be positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2

(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides

(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63

mass = 63 / 11.5

Mass = 5.48 kg

(b)
Since the elevator and lamp are going upward, I choose upward forces to be positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7

-53.70 + T = 5.48 * 1.7

T = 53.70 + 9.316 = approx 63 N

The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated upward at the same rate of 1.7 m/s^2

3 0

3 years ago