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svetoff [14.1K]
3 years ago
13

What does the symbol 'Mg' represent?​

Physics
2 answers:
RSB [31]3 years ago
3 0

Answer:

the \: symbol \: mg \: represent \: magnesium \\ thank \: you

frutty [35]3 years ago
3 0

Answer:

Magnesium.

I hope it's helps you

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Before beginning an experiment Mrs. Wilson warns her science class not to drop the bar magnets on the floor what is the most lik
Ainat [17]

Answer:

They could lose their magnetism if they are dropped too hard or banging against something (the ground)

Explanation:

the domains may bump out of alignment. hope this helps ! :)

7 0
3 years ago
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track an
pochemuha

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

8 0
4 years ago
Steam machines are used to sterilise kitchen work surfaces.
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Answer:

Styleusso customer care number 8388026087//8388026087 refund ke liye call kare

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3 years ago
The heat felt when putting a hand near a light bulb this is an example of conduction, convection, radiation?
Nesterboy [21]

Answer:

conduction

Explanation:

4 0
3 years ago
Read 2 more answers
A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel
marusya05 [52]

(a) 7.32\cdot 10^7 Hz

The frequency of an electromagnetic waves is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda=4.1 m is the wavelength of the wave in the problem

Substituting into the equation, we find

f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz

(b) 4.60\cdot 10^8 rad/s

The angular frequency of a wave is given by

\omega = 2\pi f

where

f is the frequency

For this wave,

f=7.32\cdot 10^7 Hz

So the angular frequency is

\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s

(c) 1.53 m^{-1}

The angular wave number of a wave is given by

k=\frac{2\pi}{\lambda}

where

\lambda is the wavelength of the wave

For this wave, we have

\lambda=4.1 m

so the angular wave number is

k=\frac{2\pi}{4.1 m}=1.53 m^{-1}

(d) 1.03\cdot 10^{-6}T

For an electromagnetic wave,

E=cB

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

I=\frac{E^2}{2\mu_0 c}

where we have

E = 310 V/m is the amplitude of the electric field

\mu_0 is the vacuum permeability

c is the speed of light

Substituting into the formula,

I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2

(g) 1.53\cdot 10^{-8} kg m/s

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

\frac{dp}{dt}=\frac{A}{c}

where the <S> is the magnitude of the Poynting vector, given by

=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2

and where the surface is

A = 1.8 m^2

Substituting, we find

\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s

(h) 8.47\cdot 10^{-7} N/m^2

For a surface that totally absorbs the wave, the radiation pressure is given by

p=\frac{}{c}

where we have

=254.2 W/m^2

c=3\cdot 10^8 m/s

Substituting, we find

p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2

8 0
3 years ago
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