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elena-14-01-66 [18.8K]

3 years ago

10

In the diagram, q1 = -6.3910^-9 C and q2 = +3.2210^-9 C. What is the electric field at point P? Include a + or - sign to indic

ate the direction.

1 answer:

AleksAgata [21]3 years ago

6 0

Correct Answer:

-391.1077613

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Please help, thanks!

statuscvo [17]

Because it demonstrates the relationship between a body and the forces acting upon it, and its motion in response to those forces. [Hope that helps]

7 0

2 years ago

Help me with this review question please.

QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

6 0

2 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

2 years ago

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0

3 years ago

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

kirill115 [55]

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

8 0

2 years ago