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IrinaVladis [17]
3 years ago
11

What the density of an object that has a volume of 20ml and mass of 3g.​

Physics
1 answer:
Oxana [17]3 years ago
8 0

Answer:

The equation of D = m/V

Where D = density

m = mass

and V = volume

We are solving for V, so with the manipulation of variables we multiply V on both sides giving us  

V(D) = m  

now we divide D on both sides giving us

V = m/D  

We know our mass which is 600g and our density is 3.00 g/cm^3

so

V = 600g/3.00g/cm^3 = 200cm^3  or 200mL

a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL

 

If you wish to change it to L, you'd have to convert

Explanation:

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In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

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spectrum

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