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AVprozaik [17]
3 years ago
13

When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con

sequence of rubbing the rod with the fur,
A. the rod and fur both gain mass.
B. the rod and fur both lose mass.
C. the rod gains mass and the fur loses mass.
D. the rod loses mass and the fur gains mass.
E. none of the above
Physics
1 answer:
Yuri [45]3 years ago
4 0

Answer: C. the rod gains mass and the fur loses mass.

Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.

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In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir
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Answer:

1. The nucleus is about 1/2 the size of the atom

Explanation:

Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size

Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles

so correct answer will be

1. The nucleus is about 1/2 the size of the atom

5 0
3 years ago
Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

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What category of the electromagnetic spectrum has a wavepength of 108m​
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Answer: Radio Waves

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A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th
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Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

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heat transferred to the system, q₁  30 kJ

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heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ =  35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

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