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Sergeu [11.5K]
3 years ago
7

Charles law increases keep pressure constant then you observe blank

Physics
1 answer:
OLga [1]3 years ago
7 0

If you decrease the pressure of a fixed amount of gas, its volume will increase.

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barxatty [35]

Answer:

NE DIYON INGILIZ MISIN SEN

7 0
3 years ago
What would happen if the distance between the earth and the moon decreased
makkiz [27]
The gravitational force between earth and moon will increase.
by F = GMm/r^2
F is inversely proportional to r^2
when r decrease, F will increase.
6 0
3 years ago
Read 2 more answers
Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche
DaniilM [7]

Answer:

R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}

Explanation:

Using kinematics equations:

\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t

Use \Delta y = 0 due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

t=\frac{2v_{0y}}{g}

Use the expression in the first equation to have

R = \frac{2v^2 \cos\theta\sin\theta}{g}

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

5 0
3 years ago
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
A turtle moves at an average speed of 0.2m/S towards the finish line that is 80m away from the starting line. A rabbit,racing wi
klasskru [66]

Answer:

5.6mls

Explanation:

3 0
3 years ago
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