All categories

3 years ago

Which technology contains at least one permanent magnet?

Physics

2 answers:

stich3 [128]3 years ago

6 0

I’m pretty sure it’s, B. An Electric Motor. I apologize in advance if it’s incorrect.

Xelga [282]3 years ago

4 0

Answer: B

Explanation: took it on ape x I know the other guy said it this is just for reassurance

You might be interested in

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

Which of the following terms corresponds to #2 on the image?

Mila [183]

Answer:

Trough

Explanation:

cuz physics you see

7 0

3 years ago

Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.

otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, $v_y$ ≈ 7.878 m/s.

6 0

3 years ago

A magnetic compass is placed near an insulated copper wire. When the wire is connected to a battery and a current is created, th

Luba_88 [7]

4

Every current through a wire produced a magnetic field. And since the magnetic field of Earth is weak, it will get attracted towards the wire.

4 0

3 years ago

Read 2 more answers

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

3 years ago