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3 years ago

In the equation F = Kq1 q2/r2 solve for q2. Solve for r.

Physics

1 answer:

Tpy6a [65]3 years ago

3 0

Answer:

r = √(k q₁ q₂ / F)

Explanation:

F = k q₁ q₂ / r²

Multiply both sides by r²:

F r² = k q₁ q₂

Divide both sides by F:

r² = k q₁ q₂ / F

Take the square root of both sides:

r = √(k q₁ q₂ / F)

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Which of the following air conditions would be least likely to have precipitation 10 pts

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Answer:

D. Cold, dry air

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3 years ago

.89m/s A vehicle that starts to move from the rest gets an acceleration of 5m/s² within 2 seconds. Calculate the velocity and di

GuDViN [60]

<u>Explanation:</u>

s = ? u = 0m/s v = ? a = 5m/s² t = 2s

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2 years ago

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01x10⁵ Pa.

Cerrena [4.2K]

Answer:

$\frac{1.019}{1}$

Explanation:

To solve this equation we will have to consider that the bubble is filled with an Ideal Gas and as such we can use the Ideal Gas Law

$PV = nRT$

Where

$P$ = Pressure

$V$ = Volume

$n$ = Moles

$R$ = Ideal Gas Constant

$T$ = Temperature

Now since we know that the value for the temperature and moles is constant we can simply use Boyles Law for the two states

$P_{1} V_{1} =P_{2} V_{2}$

Let us look at the two states

State 1 (at top)

Pressure = $1.01*10^5$

Volume = $V_{1}$

State 2 (at bottom)

Pressure = $1.01*10^5 + dgh$

Where

$d$ = Density of liquid (1000 kg/m³)

$d$ = Acceleration due to gravity (9.8 m/s²)

$d$ = Height of liquid (0.200 m)

Pressure = $102,962$

Volume = $V_{2}$

Inputting these values into the Boyles Law

$P_{1} V_{1} =P_{2} V_{2}\\ (101000)V_{1} = (102962)V_{2}\\ \frac{V_{1}}{V_{2}} = \frac{102962}{101000} \\ \frac{V_{1}}{V_{2}} = \frac{1.019}{1}$

6 0

3 years ago

In a chemical equation, where do the products appear

tester [92]

Answer:

products would appear after the raw materials

Explanation:

raw material + raw material = product (anything deriving from combining two materials)

4 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago