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nlexa [21]
3 years ago
6

A fire breaks out and increases the Kelvin temperature of a cylinder of compressed gas by a factor of 2.4. What is the final pre

ssure of the gas relative to its initial pressure?
Physics
1 answer:
kherson [118]3 years ago
7 0

Answer:

P(final) is 2.4 times P(initial).

Explanation:

Here we can assume that the cylinder did not break and it's volume and number of moles of gas present in the cylinder remains constant.

Given the temperature increases by a factor of 2.4. Let us assume that the initial temperature be T_{1} and the final temperature be T_{2}.

Given that T_{2}=2.4\times T_{1}

Now we know the ideal gas equation is PV=nRT

here V=constant , n=constant , R=gas constant(which is constant).

\frac{P}{T}=constant

\frac{P_{1} }{T_{1}}=\frac{P_{2} }{T_{2} }

P_{2}=(\frac{T_{2} }{T_{1} } )P_{1}

P_{2}=(\frac{2.4T_{1} }{T_{1} } )P_{1}

P_{2}=2.4\times P_{1}

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Examine the scenario.
vovangra [49]

Answer:

acceleration 8 km/h/s south

Explanation:

First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.

Based on this definition, we can already rule out the following two choices:

distance: 40 km

speed: 40 km/h

Since they only have magnitude, they are not vectors.

Then, the following option:

velocity: 5 km/h north

is wrong, because the car is moving south, not north.

So, the correct choice is

acceleration 8 km/h/s south

In fact, the acceleration can be calculated as

a=\frac{v-u}{t}

where

v = 40 km/h is the final velocity

u = 0 is the initial velocity

t = 5 s is the time

Substituting,

a=\frac{40 km/h-0}{5 s}=8 km/h/s

And since the sign is positive, the direction is the same as the velocity (south).

7 0
3 years ago
Identify the forces acting on the cars and explain why the cars do not slide down the hill
Dvinal [7]
<span>Examples of outside forces acting on a car is gravity, wind, and other cars. Cars do not slide down hills because their weight, combined with the friction of their tires against the road, hold them in place. </span>
5 0
3 years ago
A girl is standing 150m in front of a tall building, fires a shot with starting pistol. A boy standing 350m behind her, hears tw
Natasha2012 [34]

Answer:

300 m/s

Explanation:

The difference in time between the two bangs is 1 s.

Thus;

t2 - t1 = 1

We know that distance/time = speed.

Thus;

d2/v - d1/v = 1

Multiply through by v to get;

d2 - d1 = v

Where v is speed of sound in air.

d1 = 350 m

d2 = (150 × 2) + 350 = 650 m

Thus;

v = d2 - d1 = 650 - 350 = 300 m/s

8 0
3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
What is the relationship between the normal force and weight
GalinKa [24]

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

4 0
3 years ago
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