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Pavlova-9 [17]
3 years ago
13

........................

Physics
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

14°F = 263.15 K

31°F = 272.594 K

71°F = 294.817 K

35°F = 274.817 K

Explanation:

Use the formula 273.15 + ((°F–32)•5)/9

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A charge Q is transferred from an initially uncharged plastic ball to an identical ball 17 cm away. The force of attraction is t
Alecsey [184]

Answer:

2.92 x 10¹² electrons

Explanation:

given,

distance between two plastic ball, r = 17 cm

                                      r = 0.17 m

Force of attraction = F = 68 mN

                F = 0.068 N

number of electron transferred from one ball to another.

using Coulomb Force equation

F = \dfrac{kq^2}{r^2}

0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}

q² = 2.1835 x 10⁻¹³

q = 4.67 x 10⁻⁷ C

now, number of electron

 N = \dfrac{q}{e}

e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C

  N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}

       N = 2.92 x 10¹² electrons

electrons were transferred from one ball to the other is 2.92 x 10¹²

5 0
4 years ago
Why could the beam of particles not have a neutral charge?
tino4ka555 [31]

Answer:

The positive and negative charges balance each other. Overall, the atom is uncharged (neutral). However, if something happens to make an atom lose or gain an electron then the atom will no longer be neutral. An atom that gains or loses an electron becomes an ion.

Explanation:

I think this is the answer : D

4 0
3 years ago
Light wands are small tubes that can be used to give off light without electricity or fire.
miv72 [106K]
<span>c.the chemical energy of the fluids inside the wand</span>
4 0
3 years ago
Read 2 more answers
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090
cupoosta [38]

Answer:209.98 kJ

Explanation:

mass of water m=456 gm

Initial Temperature of Water T_i=25^{\circ}C

Final Temperature of water T_f=-10^{\circ}C

specific heat of ice c=2090 J/kg-K

Latent heat L=33.5\times 10^4 J/kg

specific heat of water c_{water}=4.184 KJ/kg-K

Heat require to convert water at T=25^{\circ}C to T=0^{\circ}C

Q_1=0.456\times 4.184\times (25-0)=47.69 kJ

Heat require to convert water at T=0^{\circ} to ice at T=0^{\circ}

Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ

heat require to convert ice at T=0^{\circ} C\ to\ T=-10^{\circ} C

Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ

Total heat Q=Q_1+Q_2+Q_3

Q=47.69+152.76+9.53=209.98 kJ

7 0
3 years ago
How much heat is absorbed by 60 g of copper when it is heated from 20°C to 80°C?
ira [324]

Answer:

1,836J

Explanation:

6 0
4 years ago
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