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CaHeK987 [17]
1 year ago
12

The density of a gas sample in a balloon is 1.50 g/l at 75°c. what is the density of this gas when the temperature is changed t

o 25°c at constant pressure?
Physics
1 answer:
maksim [4K]1 year ago
8 0

At 25°C, a gas sample's density in the balloon is 4.500 g/L.

When density decreases, temperature increases. When more temperature increases, density reduces. When the temperature decrease, density increases.

When a substance is heated, the molecules move faster and slightly farther apart, taking up more space and causing the density to drop. When something is cooled, the molecules slow down and get a little closer together, taking up less space and becoming denser.

The density of a gas sample in a balloon = 1.50 g/L

Temperature = 75°C

Temperature increased = 25°C

The density of a gas sample in the balloon at 25°C = (1.50 × 75)/ 25

                                                                                      = 4.500 g/L

Therefore, the density of a gas is 4.500 g/L.

Learn more about density here:

brainly.com/question/6838128

#SPJ4

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1) Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way ba
timurjin [86]

Answer:

<h3>B. 19miles</h3>

Explanation:

If Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way back home he out  of gas 6 miles after leaving the supermarket, the distance travel by fred will be the sup of all the distances he covered throughout the journey.

Distance covered by fred = 4miles + 9miles + 6miles

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8 0
3 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

5 0
2 years ago
Read 2 more answers
6) Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and blac
KIM [24]

Answer:

A) 1.5 v

B) Top plate is at higher voltage than the bottom plate

Explanation:

Battery value set between 0.0 V and 1.5 V

a) The potential difference between the plates

Δ V = V1( potential at top plate) - V2( potential at lower plate )

potential at top plate = 1.5 V

potential at lower plate = 0.0 V

hence potential difference = 1.5 V

b ) The top plate is always connected to the positive terminal of the DC source ( which is at a higher potential )while the bottom plate is connected to the negative terminal of the DC source ( which is at a lower potential )

hence the Top plate is at higher voltage than the bottom plate

7 0
3 years ago
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