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4 years ago

If gas in a sealed container has a pressure of 50 kPa at 300 K, what will the pressure be if the temperature rises to 360 K?

1 answer:

8 0

Since gas is sealed in container, volume of gas will ramain constant. from ideal gas equation PV=nRT, P/T = nR/V ==> P/T = constant.( constant since v is constant so whole term must be constant). therefore, (P/T)1 = (P/T)2.==> 50/300 = P2/360. ==> P2= 50X360/300 = 60KPa. Hope this helps

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How long will it take to fall 50 m?

Ad libitum [116K]

Answer:

down below

Explanation:

Since we aren't the given the time, lets say that an object to 25 seconds to fall 50 meters. We can use the formula [ s = d/t ] to solve.

s = 50/25

s = 2

Therefore, the object was falling at a rate of 2 meters per second.

Best of Luck!

6 0

3 years ago

charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

3 years ago

Which statement correctly compares ultraviolet light to visible light? Ultraviolet light has both a lower frequency and longer w

Effectus [21]

The correct statement is

Ultraviolet light has both a higher frequency and a higher radiant energy than visible light.

because ultraviolet light has wavelength smaller than the visible light hence has a greater frequency as compared to visible light. (frequency is inversely related to wavelength. hence smaller the wavelength , greater will be the frequency)

we also know that the radiant energy is directly proportional to the frequency. hence greater the frequency , greater will be the radiant energy.

Since the frequency is greater for ultraviolet light , it radiant energy is also greater

7 0

3 years ago

Read 2 more answers

What is one way radiation is used that is beneficial for our health?

xenn [34]

Answer:in X-rays

Explanation:

Radiation are used in X-rays for the treatment of tumours, and some skin diseases,cancer cells can be destroyed by X-rays

3 0

3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago