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larisa [96]
3 years ago
6

Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?

Physics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

           P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

            i = i₁ + i₂

if the two light bulbs are the same

           i = 2 i₁

           i₁ = i / 2

so the power of each bulb is is

           P = i₁² R

           P = R i² / 4

           P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

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Answer:

a)V=\dfrac{5.3}{P}

b)ML^{-4}T^{-2}.

Explanation:

Given that

Boyle's law

P V = Constant ,at constant temperature

a)

Given that

P_1=50KPa

V_1=0.106m^3

We know that for PV=C

P_1V_1=P_2V_2=PV

Now by putting the values

PV= 50 x 0.106

V=\dfrac{5.3}{P}

Where P is in KPa and V is in m^3

b)

PV= C

Take ln both sides

So \ln(PV)=\ln C

lnP + lnV =lnC               ( C is constant)

By differentiating

\dfrac{dP}{P}+\dfrac{dV}{V}=0

So

\dfrac{dP}{dV}=-\dfrac{P}{V}

When P= 50 KPa

\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}

It indicates the slope of PV=C curve.

It unit is \dfrac{Pa}{m^3}.

Or we can say that ML^{-4}T^{-2}.

5 0
3 years ago
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krek1111 [17]
Mass= density x volume
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If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, A)compression B)w
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Salutations!

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy,  _______________ is being done.

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Thus, your answer is option B.

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

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The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

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