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3 years ago

Somebody correct me if I’m wrong but here’s my problem.

Physics

2 answers:

nikdorinn [45]3 years ago

7 0

From the info you provided yes..

Mashutka [201]3 years ago

4 0

Maybe but not necessarily. It could possibly bring it to a 90 or higher but it would also even out with the rest of your grades. Add them all up and divide by how many there are to determine what your average would be. If you have a school tool or schoology or something like that online, you can easily see your grades to do the math.

I hope this helped :)

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A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

bekas [8.4K]

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

6 0

3 years ago

What occurs when the moon blocks the view of the sun

Sever21 [200]

Answer:

solar eclipse

Explanation:

because at that time, the moon completely covers the path and casts its shadow on earth because it is present between sun and earth's path. so, solar eclipse occurs.

hope it helps :)

8 0

3 years ago

a car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s? for 15 s. At the end of this time,

Firlakuza [10]

V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)

V = 15 + 2 X 15

V = 45 m/s

6 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

A 357 kg box is pulled 8.00 m up a 30° frictionless, inclined plane by an external force of 4925 N that acts parallel to the pla

Sidana [21]

I'm so tremble right now

4 0

3 years ago