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dezoksy [38]
4 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]4 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]4 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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A 100 N force is applied to a box at an angle of 60° to the
Anarel [89]

Answer: lift force = 100sin60 = 86.6 N

             pull force = 100sin60 = 50.0 N

Explanation:

8 0
3 years ago
skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular
laiz [17]

Answer:

L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

L=I\omega

Substitute all the values,

L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

So, its angular momentum is equal to 11.3\ kg-m^2/s.

8 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
3 years ago
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