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dezoksy [38]
2 years ago
9

Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed

both with a piece of wool.
What did Sonia observe, and why?
Physics
2 answers:
irina1246 [14]2 years ago
6 0

Answer:

Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.

Explanation:

<u>Did it on Egde 2020</u>

lions [1.4K]2 years ago
3 0

Answer: They will repel each other.

Explanation:

Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.

Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

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Anne is working on a research project that involves the use of a centrifuge. Her samples must first experience an acceleration o
oee [108]

Answer:

The final rotation is 2.449 times the initial rotation

Explanation:

Centripetal acceleration and rotation speed of particle is related as

a=rω²    

where ω  is angular speed and r radius of circle

Rearranging equation    

ω=\sqrt{\frac{a}{r} }

The initial acceleration is given by

            a₁ =100g

The final acceleration of particle is

            a₂ = 6 (100g)

                  =600g

The final and initial rotation speed are related as

                  \frac{w1}{w2} =\sqrt{\frac{a2}{a1 }

                  ω2=\sqrt{\frac{600g}{100g} } *w1

                   ω2=2.449ω1

6 0
2 years ago
Water has a specific heat of 4.186 J/g°C, and ethanol has a specific heat of 2.450 J/g°C. Based on this information, which best
arlik [135]
I think the correct answer from the choices listed above is the second option.  Based on this information, we can say that there are more molecules in a gram of water since more energy is required to raise the temperature 1 gram of water than to raise the temperature of  1 gram of ethanol.
5 0
2 years ago
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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c
puteri [66]

Answer:

64.59kpa

Explanation:

See attachment

6 0
2 years ago
A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is
Delicious77 [7]

Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
2 years ago
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