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2 years ago

How could a lightbulb near an electromagnet, but not touching it, be lit? Is ac or dc required? Defend your answer.

Physics

1 answer:

Ira Lisetskai [31]2 years ago

5 0

Answer:

If the light bulb is connected to a wire loop that intercepts changing magnetic field lines from an electromagnet, voltage will be induced which can illuminate the bulb. Change is key, so to stay lit the electromagnet should be powered with AC.

Explanation:

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It took a squirrel 0.50\,\text s0.50s0, point, 50, start text, s, end text to run 5.0\,\text m5.0m5, point, 0, start text, m, en

STALIN [3.7K]

Answer:

-5.0m/s

Explanation:

3 0

2 years ago

Read 2 more answers

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0

2 years ago

Read 2 more answers

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

4 0

2 years ago

A signal generator has an output voltage of 2.0 V with no load. When a 600 Ω load is connected to it, the output drops to 1.0 V.

kicyunya [14]

Answer:

600 Ω

Explanation:

when there is no load attached to the generator the circuit is open and zero current flows through the circuit, hence voltage drop across the internal resister is zero.

when 600Ω load is connected current starts flowing through the circuit and some voltage will drop across the internal resister.

voltage across the load resister is 1 V, so the current through it will be:

I=V/R

I=1/600 A= 1.67mA

voltage drop in the internal resister is:

V= input voltage - output voltage

V=2 V-1 V

⇒V=1 V

Now by using ohm's law

R=V/ I

R= $\frac{1}{1.67*10^{-3} }$ (I= 1.67mA, as the resistors are connected in series)

⇒R=600 Ω

Hence Thevinin resistance of the generator is 600Ω.

5 0

2 years ago

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

$v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s$

Acceleration

$a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2$

Acceleration of the tip of the plane is 0.17724 m/s²

3 0

2 years ago