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adelina 88 [10]

2 years ago

5

Two metal balls have charges of 7.1 × 10-6 coulombs and 6.9 × 10-6 coulombs. They are 5.7 × 10-1 meters apart. What is the force

of interaction between the two balls? (k = 9.0 × 109 newton·meters2/coulomb2)

2 answers:

Murljashka [212]2 years ago

6 0

The answer is 1.4 newtons <span />

Vanyuwa [196]2 years ago

4 0

Answer : The force of interaction between them is 1.35 N

Explanation :

Charge 1 $q_1=7.1\times 10^{-6}\ C$

Charge 2, $q_2=6.9\times 10^{-6}\ C$

Distance between charges, $d=5.7\times 10^{-1}\ m$

The force of interaction between them is given by using Coulomb's law. It is written as :

$F=k\dfrac{q_1q_2}{d^2}$

k is electrostatic constant

$F=9\times 10^9\times \dfrac{7.1\times 10^{-6}\ C\times 6.9\times 10^{-6}\ C}{(5.7\times 10^{-1}\ m)^2}$

$F=1.35\ N$

Hence, this is the required solution.

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Answer:

20000 Pa

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2 years ago

A student tosses a ball horizontally from a balcony to a friend 3.8 meters down below them. How long does the ball take to reach

Vsevolod [243]

Answer:

The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds

Explanation:

The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend

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Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0

The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;

h = 1/2 × g × t²

Where;

g = The acceleration due gravity of the ball = 9.81 m/s²

t = The time of motion to cover height, h

Then height is already given as h = 3.8 m

Substituting gives;

3.8 = 1/2 × 9.81 × t²

t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²

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2 years ago

Jon's bathtub is rectangular and its base is 18 ft2. (a) How fast is the water level rising if Jon is filling the tub at a rate

Kitty [74]

Answer:

The water level in the bath tub is rising at a rate of 0.0111 ft/s

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(dh/dt) = 0.0111 ft/s

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

So

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

So

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

So

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

So

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

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2 years ago

A sound wave is passing by a student 250 times every second. What is the frequency of the sound wave?

inysia [295]

My answer i believe is simply 250 Hz, because sounds or vibrations travel in 1 cycle/second, meaning the number of cycles, in your case 250, divided by the time,1 second, will ultimately be 250 Hertz. For every Cycle/second it will equal 1 Hz, so 250/1 = 250Hz

7 0

3 years ago

Read 2 more answers