Whose law states that an induced current will create a magnetic field?

If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le

GarryVolchara [31]

Answer:

$F_{y}$ = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

B₀ - W = 0

B₀ = W

ρ_fluid g V_fluid = W

the volume of the fluid is the area of the cube times the height it is submerged

V_fluid = A y

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

B - W = m a

ρ_fluid g A (y₀ + y) - W = m a

ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

ρ_fluid g A y = m a

this equation the net force is

$F_{y}$ = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

8 0

3 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

andre [41]

Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

$\displaystyle C=\frac{\epsilon_o A}{d}$

where

$\epsilon_o=8.85\cdot 10^{-12}\ F/m$

A = area of the plates = $\pi r^2$

d = separation of the plates

$\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}$

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

$\displaystyle Z=\frac{\sqrt{2}V}{I_p}$

$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

$\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}$

$X_c=36176.34\ \Omega$

The capacitance is now

$\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F$

The radius of the plates is

$r=18\ cm/2=9 \ cm = 0.09 \ m$

The separation between the plates is

$\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}$

$d=0.00184\ m$

$\boxed{d=1.84\ mm}$

8 0

2 years ago

Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?

kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time = 4s

Momentum = 1000kgm/s

Unknown:

Force = ?

Solution:

To solve this problem, we use Newton's second law of motion;

Ft = Momentum

F is the force

t is the time

So;

F x 4 = 1000kgm/s

F = 250N

6 0

2 years ago

What is the relationship between the force of gravity between two objects in regard to their mass and the distance between them?

Mashcka [7]

Answer:

hi

Explanation:

5 0

3 years ago

True or False: Speed is a vector quantity.

Travka [436]

Answer:

it's false coz

Speed is a scalar quantity.

3 0

3 years ago

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