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3 years ago

Please help will give brainliest, i might have more later

Physics

1 answer:

aleksandr82 [10.1K]3 years ago

6 0

First pic:
A

Second pic:
B

Third pic:
D

Fourth pic:
A

If you need an explanation, feel free to ask

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How much heat is givne out when 6.0g of water freezes at 0 degrees celcius the heat of fusion of water is 80.0 cal/g?

kirza4 [7]

The heat of fusion must be approx from 0 to infinity

8 0

3 years ago

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

Makovka662 [10]

Answer:

The tube should be held vertically, perpendicular to the ground.

Explanation:

As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.

And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.

8 0

3 years ago

Alguien me puede ayudar con esa tabla

Scilla [17]

私は英語がわかりません sorry pls translate

7 0

3 years ago

A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha

Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

$PQ = \sqrt{4^{2}+3^{2}}=5 m$

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

$E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C$

Electric field at P due to the charge q is

$E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C$

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C$

3 0

3 years ago

A constant voltage is applied across a circuit element. If the resistance of the element is doubled, what is the effect on the p

miv72 [106K]

Answer:

The power dissipated is reduced by a factor of 2

Explanation:

The power dissipated by a resistor is given by:

$P=I^2 R$

where

I is the current

R is the resistance

by using Ohm's law, $I=\frac{V}{R}$ , we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

$P=(\frac{V}{R})^2R=\frac{V^2}{R}$

In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have $R'=2R$ while V remains the same; substituting into the formula, we have:

$P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}$

so, the power dissipated is reduced by a factor 2.

5 0

3 years ago