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3 years ago

The answer pleaseeeeeee

Physics

1 answer:

Anarel [89]3 years ago

7 0

I’d love to help, but I can barely understand what you put

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Earth's curvature causes _____ heating of the surface by sun.<br> (one word)

DIA [1.3K]

The one word you're looking for to fill in the blank
can be "uneven" or "non-uniform".

8 0

3 years ago

Read 2 more answers

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

3 years ago

Why is atomaspheric pressure greater at the surface on Earth

andrew-mc [135]

Hello!

Because as you get closer to the surface of the earth, the more air that is on top of you. At the top of the atmosphere, there is less air, and everything is a vacuum, where you have no weight. When you get close to the earth, the weight of the air builds until it when you're at the very lowest point of the earths surface, all the air in the atmosphere above you is pressing down.

Thank You!

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3 years ago

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Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

$E_1_x = 0V/m\\E_1_y=100V/m$

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

$E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m$

∑x: $E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m$

∑y: $E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m$

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

$||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees$

or 27.23° to the left of E1.

8 0

3 years ago

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

Bumek [7]

Explanation:

Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.

Hence, $\Delta E$ = 0. And, as $\Delta E$ = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

Q = -(19000 J + 2000 J)

= -21000 J

or, |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

5 0

3 years ago