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3 years ago

A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?

Physics

2 answers:

Sidana [21]3 years ago

8 0

This question is about as sneaky as they ever get.

First let's do the easy part:

Power = (voltage) x (current)

150 watts = (120 volts) x (current)

current = (150 watts) / (120 volts)

current = 1.25 Amperes but this is NOT the answer to the question.

The voltage at the outlet is a "sinusoidal" wave ... it wiggles up and down 60 times every second. The number of "120 volts" is NOT the "peak" of the wave. In fact , the highest it ever gets is √2 greater than 120 volts. And all of this applies to the current too.

The RMS current through the lamp is (150/120) = 1.25 Amperes .

The peak current through the lamp is 1.25·√2 = about 1.77 Amperes .

devlian [24]3 years ago

3 0

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120 Divide by 120

150/120 = I

5/4 = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by √2

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Answer:

9.9 seconds

Explanation:

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3 years ago

Many students use earbuds or headphones while working on a computer or listening to music. Electromagnetics are within the liste

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2 years ago

A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above

earnstyle [38]

Answer:

a) a = 17.1 m / s², b) F = 3.04 N

Explanation:

This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities

* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components

* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components

We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is

θ = 10 -30 = -20º

The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force

sin (-20) = F_{y} / F

cos (-20) = Fₓ / F

F_{y} = F sin (-20)

Fₓ = F cos (-20)

F_y = 18 sin (-20) = -6.16 N

Fₓ = 18 cos (-20) = 16.9 N

The decomposition of the weight is the customary

sin 30 = Wₓ / W

cos 30 = W_y / W

Wₓ = W sin 30 = mg sin 30

W_y = W cos 30 = m g cos 30

Wₓ = 0.8 9.8 sin 30 = 3.92 N

W_y = 0.8 9.8 cos 30 = 6.79 N

Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane

Now we can write Newton's second law for each axis

X axis

Fₓ - fr = m a

Y Axis

N - $F_{y}$ - Wy = 0

N =F_{y} + Wy

N = 6.16 + 6.79

They both go to the negative side of the axis and

N = 12.95 N

The friction force has the formula

fr = μ N

we substitute

Fₓ - μ N = m a

a = (Fₓ - μ N) / m

we calculate

a = (16.9 - 0.25 12.95) / 0.8

a = 17.1 m / s²

b) now the block slides down with constant speed, therefore the acceleration is zero

ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced

Newton's law for the x axis

Fₓ -fr = 0

Fₓ = fr

F cos 20 = μ N

F = μ N / cos 20

we calculate

F = 0.25 12.95 / cos 20

F = 3.04 N

this is the force applied at an angle of 10º to the horizontal

3 0

3 years ago

HELP ASAP!!!

butalik [34]

Answer:

D. The acceleration after it leaves the hand is 10 m/s/s downwards

Explanation:

Vertical Throw

When an object is thrown upwards, it describes a special type of motion ruled only by gravity.

When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.

The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.

With all these concepts in mind, we state that:

D. The acceleration after it leaves the hand is 10 m/s/s downwards

The other options are not correct because:

A. The acceleration is never upwards

B. The acceleration is never 0

C. Both times are equal

8 0

3 years ago

The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

Law Incorporation [45]

The mass of the block of ice is 2.24 kg