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kiruha [24]
3 years ago
5

Which electromagnetic wave type has the largest wavelength?

Physics
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

Nearly all frequencies and wavelengths of electromagnetic radiation can be used for spectroscopy. Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation. Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.

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What part of an atom is involved in electricity and magnetism?*
-BARSIC- [3]

Answer:

Electrons

Explanation:

Electrons are very important in the world of electronics. The very small particles can stream through wires and circuits, creating currents of electricity. The electrons move from negatively charged parts to positively charged ones.

I hope this helps!

6 0
3 years ago
Read 2 more answers
For this triangle...?
Mariulka [41]
<h3><u>Answer</u>;</h3>

= 0.6

<h3><u>Explanation</u>;</h3>

Using Pythagoras theorrem

Base² + height ² = Hypotenuse²

Thus;

Base² = 15² - 12²

          = 81

Base = √81 = 9

But; cosine = adjacent/hypotenuse

Hence; cos θ = 9/15

                      <u>= 0.6 </u>

5 0
3 years ago
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How many joules are required to heat .250 kg of liquid water from 0 °C to 100 °C ?
dedylja [7]

Answer:

Specific heat of water is 4.186 J/g/C. The heat required to raise the temperature by

is

Here  is mass of water being heated,  specific heat of water and  change in temperature.

Here .

Heat energy required is

Explanation:

4 0
3 years ago
A woman on a bicycle traveling at 10 m/s on a horizontal road stops pedaling as she starts up a hill inclined at 4.0º to the hor
IrinaK [193]
The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.

m mass
v velocity
h height
g = 9.81m/s²

The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)

6 0
3 years ago
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

4 0
2 years ago
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