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Nimfa-mama [501]
3 years ago
8

What happen to the bulb when it is in series connection?

Physics
2 answers:
coldgirl [10]3 years ago
8 0
If the bulb is in series with something else, then . . .

--  The brightness of the bulb depends on the <em>other</em> device in the circuit. 

--  If the other device is designed to use <em>less power</em> than the bulb, then the
other device gets <em>more power</em> than the bulb gets.

--  If the other device is designed to use <em>more power </em>than the bulb, then the
other device gets <em>less power</em> than the bulb gets.

--  If the other device is removed from the circuit, then the bulb doesn't light at all.

This description of the often-screwy behavior of a series circuit may partly explain
why the electric service in your home is not a series circuit.




Brums [2.3K]3 years ago
3 0
Nothing special happens. It just lights up normally. A series circuit is a normal circuit, the different one is a parallel circuit.
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7.a railway truck A of a mass of 2000kg moves westwards with a velocity of 3m/s. It collides with a stationary truck B 1200kg, l
ziro4ka [17]

Answer:

9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?

6 0
3 years ago
To halt the decline in biodiversity, we must do which of the following?
emmainna [20.7K]
<span>adopt ecological conservation practices </span>
6 0
3 years ago
To understand the experiment that led to the discovery of the photoelectric effect.
andrew11 [14]

Answer:

A) Emin = eV

B) Vo = (E_light - Φ) ÷ e

Explanation:

A)

Energy of electron is the product of electron charge and the applied potential difference.

The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;

Emin = eV

B)

The maximum stopping potential energy is eVo,

The energy of the electron due to the light is E_light.

If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy

Φ = E_light - eVo

Therefore,

eVo = E_light - Φ

Vo = (E_light - Φ) ÷ e

6 0
3 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

3 0
3 years ago
Listen →
Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

                              v = u + at

                                  = 4 + (0.5 x 30)

                                = 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of  3 seconds

                                    V = 19 + (0.5 x 3)

                                       = 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

6 0
3 years ago
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