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jeka94
2 years ago
12

Sample Response: Yes, his graph is correct because it shows that as the average kinetic energy increases, so does the temperatur

e. This is called a direct relationship. What did you include in your response? Check all that apply. As average kinetic energy increases, temperature also increases. The graph shows a direct relationship between average kinetic energy and temperature.
Physics
1 answer:
shepuryov [24]2 years ago
7 0

The graph will be correct, if it shows a direct relationship between average kinetic energy and temperature of the gas molecules.

<h3>What is average kinetic energy?</h3>

The average kinetic energy of a gas molecule is the energy possesed by the gas due to its relative motion.

Average kinetic energy of gas molecules has a direct relationship with temperature of the gas molecules. As the gas temperature increases, the kinetic energy of the gas increases and consequently, the speed of the gas increases as well.

Thus, the graph will be correct, if it shows a direct relationship between average kinetic energy and temperature of the gas molecules.

Learn more about average kinetic energy here: brainly.com/question/9078768

You might be interested in
two equal and unlike parallel forces of magnitude 34N act on a rigid body,such that the moment of couple is 8.50 Nm. calculate t
goldfiish [28.3K]

The moment of a couple is Force × perpendicular distance from the arm of the line of action

so the arm of the couple= moment of couple/force=8.5/34=0.25m

the arm is 0.25m

5 0
3 years ago
Find the mean of set values 12g 9g 13g 12g 20g 17g 15g
nalin [4]

{\huge{\boxed{\mathcal{\green{Answer}}}}} \\ \frac{12 + 9 + 13 + 12 + 20 + 17 + 15}{7}  \\  =  \frac{98}{7}  \\  = 14 \\ {\huge{\boxed{\mathcal{\green{Hope \: it \: Helps}}}}}

5 0
3 years ago
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
3 years ago
Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90
11111nata11111 [884]
<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

8 0
3 years ago
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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