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kvv77 [185]
3 years ago
5

The helium-filled balloons below are carrying negative electric charges, as shown. Attach two of the balloons to the hooks below

to show what would happen if a red balloon and a blue balloon were brought near each other. Plzz help
Physics
2 answers:
Brilliant_brown [7]3 years ago
8 0

Answer:

Here you go : )

Explanation:

JulsSmile [24]3 years ago
7 0

Answer:

these are in the hooks the balloons repel each other if the hooks are isolated.

Explanation:

In electrostatics, charges of the same sign repel and of a different sign attract each other.

With this explanation we analyze the situation presented, they indicate that the balloons have a negative taste, so when these are in the hooks the balloons repel each other if the hooks are isolated.

If the hooks are connected to Earth, the poop of the balloons is neutralized, so there is neither attraction nor repulsion.

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During an isothermal process, 10 j of heat is removed from an ideal gas. What is the work done by the gas in the process?
Black_prince [1.1K]

The work done in the isothermal process is 10 joule.

We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.

ΔU = 0

Hence,

ΔU = Q - W

0 = Q - W

Q = W

It means that the heat transferred is the same as the work done.

From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.

Find more on isothermal at: brainly.com/question/17097259

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8 0
2 years ago
A 2.0-kg ball has a momentum of 25kg.m/s what is the ball's speed?
Reil [10]

We know, momentum = mass * speed

25kgm/s = 2 kg * s

s = 25/2 = 12.5 m/s

5 0
3 years ago
Read 2 more answers
Which of the following is an example of technology influencing science?
FinnZ [79.3K]

Answer:A and C

Explanation:

8 0
3 years ago
A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
charle [14.2K]

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

6 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
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