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Ksenya-84 [330]
4 years ago
5

A hollow, transparent plastic tube is placed on a horizontal surface. A wire carrying a current is wound once around the tube to

form a circular Loop in The Wire. And what what direction would a compass placed inside the tube point?
Physics
1 answer:
tekilochka [14]4 years ago
6 0

When wire is coiled on the plastic tube and current flow through that wire then the system will behave like a solenoid in which current flows through the coiled wires and then it produce magnetic field along the axis of solenoid

So here it will also produce magnetic field along the axis of solenoid and then due to this magnetic field the compass placed inside the tube will experience torque on its needle due to which it will have tendency to oriented along the direction of magnetic field.

So here we can say that compass needle will lie along the axis of the plastic tube.

here magnetic field along the axis of tube will be given same as solenoid which is given as

B = \mu_0 ni

so here direction of compass needle is axis of the tube

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A driver who does not wear a seatbelt continues to move forward with a speed of 18.0 m/s (due to inertia) until something solid
vredina [299]

Answer:

F = 2.113 x 10⁵ N

Explanation:

First we need to calculate the deceleration of the driver by using 3rd equation of motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance = 5 cm = 0.05 m

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 18 m/s

Therefore,

2a(0.05 m) = (0 m/s)² - (18 m/s)²

a = (- 324 m²/s²)/0.1 m

a = - 3240 m/s²

where, negative sign represents deceleration

From Newton's Second Law of Motion:

F = ma

F = (65 kg)(-3240 m/s²)

F = - 2.106 x 10⁵ N

So, he closest answer is:

<u>F = 2.113 x 10⁵ N</u>

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3 years ago
A bicyclist rides forward with a positive acceleration. How would her acceleration be different if she had less mass?
Angelina_Jolie [31]
Her acceleration would increase, if mass decreases than the acceleration would increase
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3 years ago
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You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t
Nataly [62]

Answer:

Approximately \rm 2.0\; V.

Approximately \rm 30 \; mA. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here R_1= 10\; \Omega and R_2= 100\; \Omega are connected in parallel. Their effective resistance would be equal to

\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega.

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be 20\; \rm mA = 0.020\; \rm A. That should also be the current through the effective \displaystyle \rm \frac{10}{11} \; \Omega resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V.

The voltage drop across the entire circuit would equal to

  • the voltage drop across the resistors, plus
  • the voltage drop across the LED.

In this case, that value would be equal to 1.83 + 0.182 \approx 2.0\; \rm V. That's the voltage that needs to be supplied to the circuit to achieve a current of 20\; \rm mA through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega.

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega.

Again, the current in a serial circuit is the same in all appliances.

\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA.

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Diffusion probably be more specific
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allochka39001 [22]
The answer is c if i have not mistaken
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