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eduard
3 years ago
14

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 50 Ω. How

much current flows through the circuit?
Physics
2 answers:
VLD [36.1K]3 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

Therefore, 0.3joule current flows through the circuit.

Hope this helped

Phantasy [73]3 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

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A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe
d1i1m1o1n [39]

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

5 0
3 years ago
Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw
jarptica [38.1K]

Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

3 0
3 years ago
The magnitude of the momentum of an object is 64 kilogram meters per second. If the magnitude of te velocity is doubled, the mag
dsp73

Answer:

128 \frac{kg*m}{s}

Explanation:

P=m*v\\if double velocity:\\P=m*2v=2m*v\\

7 0
4 years ago
If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit
Oxana [17]

Answer:

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

4 0
3 years ago
Which of the following are true? Select all that apply. The net electric field at any location inside a block of copper is zero
Agata [3.3K]

Answer:

1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.

2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.

3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.

Explanation:

1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.

2) The charged particles on a conductor in equilibrium are at rest.

6 0
3 years ago
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