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Varvara68 [4.7K]
3 years ago
13

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

Physics
2 answers:
ivanzaharov [21]3 years ago
6 0

Answer:  

Sample Answer: The lightbulb will not light because this is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

It will.

Explanation:

For a circuit to function properly, everything has to be connected. As the diagram shows that everything is connected and there are no gaps, it will light.

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An object is dropped from a height of 50 meters. Ignoring air resistance, how long doe take to hit the ground?
Arte-miy333 [17]

here given that object is dropped from height h = 50 m

So here we can say

initial speed is ZERO

acceleration is due to gravity

now in order to find time to reach the ground we can use kinematics

y = v_i*t + \frac{1}{2}at^2

now plug in all values in it

50 = 0 + \frac{1}{2} \times 9.8(t^2)

t = 3.2 s

so it will take 3.2 s to reach the ground

3 0
3 years ago
What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make
almond37 [142]

Answer:

D

Explanation:

The particles will move side to side over large areas

6 0
3 years ago
An 18-kg bicycle carrying a 62-kg girl is traveling at a speed of 7 m/s. What is the kinetic energy of the girl
Dafna11 [192]

Answer:

1960 J

Explanation:

EK = (18 kg + 62 kg)(7.0 m/s)^{2}/2 = 1960 J

6 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
HEY CAN YALL HELP ME IN DIS!!
Natasha_Volkova [10]

It might me same volume

5 0
4 years ago
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