Give two examples of gravity in the solar system

The small metal cylinder has a mass of 0.20 kgkg, the coefficient of static friction between the cylinder and the turntable is 0

deff fn [24]

Answer:

velocity of the metal cylinder = 0.343 m/s

Explanation:

Force putting the metal cylinder is given by

F = mv²/r

But this force will balance the frictional force between the metal cylinder and the turntable

The frictional force is given by

μN = μ × mg = 0.08 × 0.2 × 9.81 = 0.15696 N

r = 0.15 m, m = 0.2 kg,

F = mv²/r = 0.2 v²/(0.15) = 1.3333 v²

1.3333 v² = 0.15696

v² = 0.117

v = 0.343 m/s

8 0

3 years ago

A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 0

3 years ago

The displacement volume of an automobile engine is 167 in3. what is the displacement volume in liters?

forsale [732]

The displacement volume in liters is 2.74 liters.

<h3>What is displacement volume?</h3>

Displacement volume is the quantity of solvent that will be displaced by a specified quantity of a solid during dissolution.

It can also be defined as the volume displaced by the piston as it moves between top dead center and bottom dead center in a car engine.

<h3>Displacement volume in liters</h3>

1 liter = 61.02 in³

? = 167 in³

= 167/61.02

= 2.74 liters

Thus, the displacement volume in liters is 2.74 liters.

Learn more about displacement volume here: brainly.com/question/1945909

#SPJ1

4 0

1 year ago

why can light from the sun cause skin cancer but the light from a light bulb cannot cause skin cancer?

meriva

Answer:

Damage from UV exposure is cumulative and increases your skin cancer risk over time. While your body can repair some of the DNA damage in skin cells, it can't repair all of it. The unrepaired damage builds up over time and triggers mutations that cause skin cells to multiply rapidly. That can lead to malignant tumors.

4 0

3 years ago

Read 2 more answers