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mote1985 [20]
3 years ago
14

suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round

increase decrease or remain the same
Physics
1 answer:
Akimi4 [234]3 years ago
3 0

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

\tau = \frac{dL}{dt}

where L = angular momentum

since here torque is ZERO

0 = \frac{dL}{dt}

L = constant

so here we can write initial angular momentum of the system as

L = (I_1 + I_2)*\omega

here we know that

I_1 = moment of inertia of merry go round

I_2 = moment of inertia of child

so here we can say

(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

I_2 = mr^2

here m = mass of child

r = distance of child from center

Since child is moving inside so his distance from center is decreasing

so here moment of inertia of child is decreasing as he starts moving inside

so final angular speed of merry go round will increase as child go inside

\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}

so here as

I_2' < I_2

final angular speed will be more than initial speed as child moves inside

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The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current
irina1246 [14]

Answer:

a  average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by  P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

6 0
3 years ago
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The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec
PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

5 0
3 years ago
What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?
Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

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A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
tatuchka [14]

We have that the values for F north, F east, F up are

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From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

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Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of
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Answer:

8.35 × 10^{-3} kg

Explanation:

Remember the formula for heat:

Q=mc\Delta T

You just need to solve for m, doing so you're left with:

m=\dfrac{Q}{c\Delta T}

Replace the variables with numbers and that's it.

7 0
2 years ago
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