Which kind of wave moves back and forth along the direction of the wave?.

So, RCF, or relative centrifugal force should be equal to RCF = <img src="https://tex.z-dn.net/?f=11%2C18%2Ar%2A%28RPM%2F1000%29

zysi [14]

After plugging all the data into the equation, the result of the relative centrifugal force (RCF) is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

ac is centripetal acceleration
g is acceleration due to gravity

$RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2$

where;

r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

$RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g$

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

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6 0

1 year ago

Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20

timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0

3 years ago

Read 2 more answers

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled

Alisiya [41]

Answer: $F_2=16\times F_1$

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ is the charges of particle

r=distance between charge particle

$=\frac{kq^2}{r^2}$ when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

$F_2=\frac{k(2q)^2}{(0.5r)^2}$

$F_2=\frac{kq^2}{r^2}\times 4\times 4$

$F_2=16\times F_1$

6 0

3 years ago

A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What

iren [92.7K]

Answer:

0.0006091222 m

Explanation:

q = Charge = 42 pC

V = Voltage = 620 V

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric potential is given by (at r = R)

$V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m$

The radius of the drop is 0.0006091222 m

3 0

3 years ago

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T

larisa [96]

Answer:

a. Decreases

b. Increases

c. Remains the same

d. Increases

Explanation:

a. Capacitance is given by c= Ak€/d

where A is conductivity plate with Area

K is a constant

€ is dielectric with permittivity.

d is the distance

b. Potential difference is given by

V = Ed, since, the electric field remains the

same, the potential diterence also increases with increase in distance.

Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.

c. Electric field remains the same because charge on the

plate remains the same.

d. since electric field remains the same and capacitance decreases, the energy increases.

E= 1/2c * Q^2

7 0

3 years ago