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2 years ago

Use the circuit diagram to decide if the lightbulb will light. Justify your answer.

Physics

1 answer:

marin [14]2 years ago

6 0

Answer:

The light bulb will NOT light because the lower switch being closed makes a short circuit. The light will have a high resistance and the electricity will choose to flow the easiest way possible through the low resistance wires.

Explanation:

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Vector A and B are in the xy plane where A is 3.2m at 45° to the positive x axis and B is 2.4m at 290° to positive x axis. Deter

Anna71 [15]

hah, YOU JUST GOT

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3 0

3 years ago

When voltage is induced in a coil of wire current is?

lara31 [8.8K]

If the wire is then wound into a coil, the magnetic field is greatly intensified producing a static magnetic field around itself forming the shape of a bar magnet giving a distinct North and South pole.

5 0

3 years ago

The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}

tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

$K.E=\frac{1}{2}mv^2$

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2$

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2$

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

$\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}$

Given that, v = 2v'

$\frac{K.E}{K.E'}=\frac{1}{1}$

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0

3 years ago

The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed

padilas [110]

Answer:

Details in the explanation

Explanation:

Vertical Launch

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

$h = -16t^2 + 64t$

Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

$h = -16(1)^2 + 64(1)=64-16=48\ ft$

The object is at a height of 48 feet

At t=2 sec

$h = -16(2)^2 + 64(2)=128-64=64\ ft$

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

$h = -16(3)^2 + 64(3)=192-144=48\ ft$

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

$h = -16(4)^2 + 64(4)=256-256=0\ ft$

The object is at ground level and has returned to the launch point.

5 0

3 years ago

A wind turbine with a rotor diameter of 40 m produces 90 kW of electrical power when the wind speed is 8 m/s. The density of air

aleksley [76]

Answer:

0.233

Explanation:

Given that

Diameter of rotor, d = 40 m

Power of rotor, P = 90 kW

Speed of the wind, v = 8 m/s

Density of air, p = 1.2 kg/m³

It is a known fact that

KE = ½mv², where mass flow rate, m

m = p.A.v, where Area, A

A = πd²/4

A = (3.142 * 40²)/4

A = 3.142 * 1600/4

A = 3.142 * 400

A = 1256.8 m², substitute for A in the mass flow rate equation

m = p.A.v

m = 1.2 * 1256.8 * 8

m = 12065.28, substitute for m in the KE equation

KE = ½mv²

KE = ½ * 12065.28 * 8²

KE = 12065.28 * 32

KE = 386088.96 W or

KE = 386.1 kW

Fraction of kinetic energy converted to electric energy is

Fraction = Electric Power / Total KE

Fraction = 90 / 386.1

Fraction = 0.233

8 0

4 years ago