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UNO [17]
2 years ago
14

Which of the following should a warm up NOT include?

Physics
1 answer:
Arlecino [84]2 years ago
7 0
The answer to this is ax
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Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

3 0
2 years ago
What is the purpose of a rearview mirror in a car?
Shalnov [3]
The third choice.

The driver wants to see the object that is behind him. The light reflects off the mirror into the eyes of the driver portraying the object behind him
5 0
2 years ago
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4. Calculate the total resistance for two 180 ohm resistors connected in<br> parallel
solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0
3 years ago
A combustion reaction usually gives off heat and light is it true or false
yanalaym [24]
The answer is true I think
3 0
3 years ago
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
bogdanovich [222]

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \sqrt{2y_{o}/g}

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \sqrt{k/m}

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \sqrt{k/3} \sqrt{2y_{o}/g}

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

8 0
3 years ago
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