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3 years ago

How are electricity and magnets connected?

Physics

2 answers:

Stella [2.4K]3 years ago

3 0

Answer:

Magnets are employed to generate electricity.

Explanation:

Magnets' characteristics are employed to generate electricity. Electrons are pulled and pushed by moving magnetic fields. When you move a magnet around a coil of wire, or a coil of wire around a magnet, the electrons in the wire are pushed out and an electrical current is created.

aleksley [76]3 years ago

3 0

Answer:

they are closely related, because flowing electrons produce some magnetic field and spinning magnets make a flow of electricity, make them interact and it's called electromagnetism

Explanation:

up there↑

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A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an a

kondor19780726 [428]

RESULT: Twalk= 3.64 hr

7 0

3 years ago

The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.

klio [65]

The relationship between frequency and wavelength for an electromagnetic wave is
$c=f \lambda$
where
f is the frequency
$\lambda$ is the wavelength
$c=3 \cdot 10^8 m/s$ is the speed of light.

For the light in our problem, the frequency is $f=1.20 \cdot 10^{13} s^{-1}$ , so its wavelength is (re-arranging the previous formula)
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}= 2.5 \cdot 10^{-5}m$

8 0

3 years ago

Add the vectors:

Anettt [7]

Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

$y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m$

and vector 2 has

$x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m$

$y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m$

Add these vectors to get the resultant, which has components

$x_{\rm total}\approx11.133\,\mathrm m$

$y_{\rm total}\approx13.268\,\mathrm m$

The magnitude of the resultant is

$\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m$

with direction $\theta$ such that

$\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ$

or about 50º N of E.

8 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

3 years ago