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3 years ago

Calculate the electric current that flows through a electric heater of power 1000watt when collected with 220 volt ac supply

Physics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

4.545 A

Explanation:

From the question,

Using,

P = VI....................... Equation 1

Where P = power of the heater, V = Voltage supply, I = Electric current that flows through the heater

make I the subject of the equation above

I = P/V.................. Equation 2

Given: P = 1000 watt, V = 220 volt.

Substitute these values into equation 2

I = 1000/220

I = 4.545 A

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How does light behave when a laser is pointed at a prism?

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2 years ago

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic

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Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

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where

m is the mass of the electron

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$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$

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3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

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3 years ago

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

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Answer:

$Fr=12544N$

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation $Fr=usmg$ , where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

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3 years ago