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pishuonlain [190]
3 years ago
10

What is the resistance of a bulb of 4owconnected in a line of 220v?2​

Physics
1 answer:
inn [45]3 years ago
8 0

Answer:

1210 ohm

Explanation:

Given :

P=40 W

V=220 V

Now,

P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm

Therefore, resistance of bulb will be 1210 ohm

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34kurt

Answer:

Please find the answer in the explanation.

Explanation:

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3 years ago
A car on a roller coaster starts at zero speed at an elevation above the ground of 26 m. It coasts down a slope, and then climbs
nirvana33 [79]

The speed of the car at the top of the hill is 14m/s

<u>Explanation:</u>

given that

Initial velocity u of the car=0 m/s

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elevation of the first slope=26 m

elevation of second slope=16m

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speed of the car at the top of the hill can be determined by using the equation

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speed of the car at the top of the hill is 14m/s

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3 years ago
Why is photosynthesis essential for plant homeostasis?
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It should be answer choice b
6 0
3 years ago
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3 years ago
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In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient
morpeh [17]

Answer:

v(t) = 21.3t

v(t) = 5.3t

v(t) = 48 -48 e ^{ \frac{t}{9}}

Explanation:

When no sliding friction and no air resistance occurs:

m\frac{dv}{dt} = mgsin \theta

where;

\frac{dv}{dt} = gsin \theta , 0 < \theta <  \frac{ \pi}{2}

Taking m = 3 ; the differential equation is:

3 \frac{dv}{dt}= 128*\frac{1}{2}

3 \frac{dv}{dt}= 64

\frac{dv}{dt}= 21.3

By Integration;

v(t) = 21.3 t + C

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta

Taking m =3 ; the differential equation is;

3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}

\frac{dv}{dt}= 5.3

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv

The differential equation is :

= 3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v

= 3 \frac{dv}{dt}=16 -\frac{1}{3}v

By integration

v(t) = 48 + Ce ^{\frac{t}{9}

Since; V(0) = 0 ; Then C = -48

v(t) = 48 -48 e ^{ \frac{t}{9}}

7 0
3 years ago
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