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Alexxandr [17]
3 years ago
14

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

al force between them
Physics
1 answer:
RUDIKE [14]3 years ago
4 0

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

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The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door,
ValentinkaMS [17]

Answer:

T0=390 degF

Explanation:

The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door, an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.

If T is temperature, then by Newton’s law

dTdt=-k(T−T0)

Where

T0− is temperature of oven  

∫\frac{dT}{T-T0} =\int\limits^0.5_0 {-k} \, dt

ln(T-T0), from 110 to 70=-kt, from 1/2 to 0

ln(110-T0)-ln(70-T0)=-k(1/2-0)

2in\frac{110-T0}{70-T0} =-k\\k=-2ln\frac{T0-110}{T0-70} ...........................1

integrating the LHS of the equation from 110 to 145F

∫∫∫\frac{dT}{T-T0} =\int\limits^1_0.5 {-k} \, dt

ln(145-T0)-ln(110-T0)=-k(1-1/2)

2ln145-T0/(110-T0)=-k

k=-2lnT0-145/(T0-110).......................2

couplijng equatiuon 2 with 1

(T0-110)^2=(T0-70)(T0-145)

T0^2-220T0+12100=T0^2-215T0+10150

5T0=1950

T0=390 degF

5 0
3 years ago
You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera
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Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

  • Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁  = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = \frac{2*4186*5}{0.5*25}= 3348.8J/kg C≅ 3349J/kg C

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lara31 [8.8K]

Answer:

potential energy or stored energy

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If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is
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The first thing you should know to solve this problem is the conversion of pounds to kilograms:
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 her mass expressed in kilograms is 56.25 Kg.
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poizon [28]

Answer:

v = 10 m/s

Explanation:

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