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mina [271]
3 years ago
5

The invisible line that passes through the North Pole and the South Pole is called Earth's what

Physics
1 answer:
REY [17]3 years ago
5 0

Answer:

The Prime Meridian

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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
A 10n falling object encounters 4n of air resistance. what is the net force on the object?
Margaret [11]
It would be 6n down.
5 0
3 years ago
Why the reflected rays for a rough surface do not result in the formation of an image.
Oksana_A [137]

Answer:

this is because the light rays get reflected irregularly

Explanation:

8 0
2 years ago
If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?
Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

W=F\times x

W=mg\times x

Where, x = distance

mg = weight

Put the value into the formula

W=40\times2

W= 80\ Nm

W=80\ J

Hence, The work done on the box is 80 J.

5 0
3 years ago
As shown in the diagram below, a 1 kg rock tied to a rope is
Tatiana [17]

Answer:

Explanation:

Angular momentum has a formula of L = mvr. Fillingin:

L = (1.0)(5.0)(1.0)

L = 5.0 kg*m/s

4 0
3 years ago
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