A complex sound’s quality depends on the mix of harmonics or overtones added to the _____________

Which factors affect the strength of the electric force between two objects

iris [78.8K]

By definition, the electric force is given by:

$f = k * \frac{q1q2}{d ^ 2}$

Where,

q1: electric charge of object number 1.

q2: electric charge of object number 2.

d: distance between both objects

k: proportionality constant

Therefore, the magnitude of the electric force is affected by:

1) The product of the charges of the objects

2) The distance between objects

Answer:

The factors that affect strength of the electric force between two objects are:

1) The product of the charges of the objects

2) The distance between objects

5 0

4 years ago

Read 2 more answers

If heat is gained from a source, it is __________ from the source.

Strike441 [17]

Answer:

Derived, or expended

4 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago

A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0

4 years ago

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$ ,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0

3 years ago

A complex sound’s quality depends on the mix of harmonics or overtones added to the ______________ frequency