All categories

2 years ago

what would happen when an inflated balloon is place in an ice bath and the atmosphere pressure has not changed?

Physics

1 answer:

ruslelena [56]2 years ago

7 0

The inflated balloon shrinks when it is placed in an ice bath with no change in atmospheric pressure.

Explanation:

When the inflated balloon is subjected to an ice bath, it shrinks. This is due to the fact that smaller volume gets occupied by the air/gas inside the balloon as the temperature decreases. Hence, causes the balloon walls to collapse.

An ice bath also lowers the overall air temperature of the balloon inside. As the temperature decreases, the air molecules move more slowly and with lower energy. Because of the particle's lower energy, their collisions with the walls are not enough to keep the inflated balloon.

You might be interested in

Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an un

Lena [83]

Answer: $316.8V$

Explanation:

given data:

metre moving current = $0.96mA$

meters voltage = $288V$

or $0.96*300V = 288V$

Solution:

$v1 = (0.96mA*150)$

$= 144V$

$i1 = \frac{144v}{750}$

$= 0.192mA$

$i2 = imovement + i1$

$i2= 0.96mA+0.192mA$

$= 1.152mA$

$Vmeasured = 144V+(150)(1.152mA)$

$=316.8V$

the unknown voltage is 316.8V

7 0

3 years ago

What is the gravitational acceleration close to the surface of a planet with a mass of 9ME and radius of 3RE, where ME and RE ar

Papessa [141]

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

5 0

3 years ago

Which biome has the most variable year round temperature

skad [1K]

That should be the Grasslands ?

5 0

2 years ago

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

2 years ago

A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0

2 years ago