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Mrac [35]
3 years ago
13

Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom

e 2q and the distance becomes 3r, what are the new mutual forces
Physics
1 answer:
jeka57 [31]3 years ago
3 0

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

You might be interested in
If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23
Bingel [31]

The complete question is :

A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .

If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?

Solution :

Given volume of the hot air balloon $=2100 \ m^3$

The balloon fabric weights = 860 N

The weight of the basket with the gear and propane tank = 1300 N

Density of outside air $= 1.23 \ kg/m^3$

∴  Total pay load = Weight of the air displaced - weight of gas inside the balloon

Total pay load = 860 + 1300 + 3400

                        = 5560 N

Mass = density x volume

Weight  $= \text{mass} \times g$

Weight = volume x density $\times \text{ acceleration due to gravity (g)}$

Weight of the displaced air = 2100 x 1.23 x 9.8

                                             = 25313 N

Weight of the gas inside the balloon = density $\times \text{ acceleration due to gravity (g)}$ x volume

                                                             = density x 9.8 x 2100

                                                             = density x 20580 N

Therefore substituting the values, we get

⇒ 25313 - (density x 20580) = 5560

⇒ density $=\frac{19753}{20580}$

                 $= 0.96 \ kg/m^3$

So the density of the heated gas $= 0.96 \ kg/m^3$

8 0
4 years ago
You are asked to build an LC circuit that oscillates at 13 kHz. In addition, you are told that the maximum current in the circui
inessss [21]

Answer:

a)L=0.00142H

b) C=2.65*10^{-12}

Explanation:

From the question we are told that:

FrequencyF=13kHz

Current I=0.14A

CapacitorC_e=1.4*10^{-5}J

Generally the equation for Energy in the inductor is mathematically given by

Where L is now subject

L=\frac{2C_e}{I^2}

L=\frac{2*1.4*10^{-5}}{(0.14)^2}

L=0.00142H

Generally the equation for Value of Capacitor is mathematically given by

C=\frac{1}{(2 \pi f)^2} L

C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142

C=2.65*10^{-12}

3 0
3 years ago
An object is falling downward at a rate of 25 m/s. Two seconds later, what is its acceleration?
zhenek [66]

acceleration times time falling

25*2

50 m*s^-1

6 0
4 years ago
A quarter placed on a turntable has a centripetal acceleration of
icang [17]

Answer:

200

Explanation:

Because I know

4 0
3 years ago
3. An engine’s fuel is heated to 2,000 K and the surrounding air is 300 K. Calculate the ideal efficiency of the engine. Hint: T
maksim [4K]

Answer: E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

Explanation:

The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH

E = W/QH.

W=QH – QC,

Where Qc is the output heat.

That is,

E=1 - Qc/QH

E =1 - Tc/TH

where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.

Note: The unit of temperature must be in Kelvin.

Tc = 300K

TH = 2000K

Substituting the values of E, we have;

E = 1 - 300K/2000K

E = 1 - 0.15

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

4 0
4 years ago
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