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OlgaM077 [116]
3 years ago
10

At one atmosphere of pressure and 25°C, a hot air balloon has a volume of 4,000 liters. While still tied to the ground, the air

in the balloon is heated to 68°C. When the balloon is untied, ready to float away, it has a new volume of
A) 1088 liters
B) 3496 liters
C) 4577 liters
D) 9280 liters
Physics
2 answers:
spin [16.1K]3 years ago
5 0

Answer:

C) 4577 Liters

Explanation:

As we know that pressure inside the balloon will remain constant so here by ideal gas equation we can say

PV = nRT

now since pressure is constant so we will have

\frac{V}{T} = \frac{nR}{P}

now for two different temperatures we will have

\frac{V_1}{T_1} = \frac{V_2}{T_2}

given that

T_1 = 25^0C = 25 + 273 = 298 K

V_1 = 4000 Ltr

T_2 = 68^0 C = 68 + 273 = 341 K

now from above equation we will have

\frac{4000}{298} = \frac{V_2}{341}

V_2 = 4577 Liters

yawa3891 [41]3 years ago
3 0

Based on the options given, the most likely answer to this query is C) 4577 liters.

Upon computation of the given variables the result seems to be 4577 L

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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At 15°C air is transmitted <br>at 340 m/s. Express this speed<br>in Kilometers per hour.​
EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s =  1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1  km/hr

1m/s = 3600/1000  km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0
3 years ago
Read 2 more answers
J.J. Thomson theorized that, if an atom had all of its negatively charged
Bumek [7]

Answer:

C. Positively charged

Explanation:

The plum pudding model of the atom proposes by J. J. Thomson consisted of electrons which lay embedded as the raisins within a dough or soup that was positively charged. The electron was discovered by J. J. Thomson in 1897 through cathode ray tube experiments.

Based on the plum pudding model, if all the negatively charge electrons contained in an atom are removed, the material remaining will be the <em>positively charged</em> soup

6 0
3 years ago
If a ball is tossed straight up into the air, at what position is its potential energy the greatest?
krok68 [10]
A. when it reaches the top of its flight

potential energy increases as the height of the object does, and the highest point of a ball's flight is when it reaches the top. 
4 0
3 years ago
Read 2 more answers
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha
ikadub [295]

We have that the spring constant is mathematically given as

k=2.37*10^{11}N/m

Generally, the equation for angular velocity is mathematically given by

\omega=\sqrt{k}{m}

Where

k=spring constant

And

\omega =\frac{2\pi}{T}

Therefore

\frac{2\pi}{T}=\sqrt{k}{n}

Hence giving spring constant k

k=m((\frac{2 \pi}{T})^2

Generally

Mass of earth m=5.97*10^{24}

Period for on complete resolution of Earth around the Sun

T=365 days

T=365*24*3600

Therefore

k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2

k=2.37*10^{11}N/m

In conclusion

The effective spring constant of this simple harmonic motion is

k=2.37*10^{11}N/m

For more information on this visit

brainly.com/question/14159361

8 0
3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
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