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DiKsa [7]
3 years ago
14

Which of the following are found within the electromagnetic spectrum? Check all that apply. sound waves visible light X rays ult

raviolet radiation gamma rays microwave radiation
Physics
1 answer:
sattari [20]3 years ago
8 0

Answer:

Visible light

X rays

ultraviolet radiation

gamma rays

microwave radiation

Explanation:

Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves have all same speed in a vacuum (c=3.0\cdot 10^8 m/s, known as speed of light) and are classified into 7 different types according to their frequency and wavelength. This classification is called electromagnetic spectrum.

From lowest to highest wavelength, the 7 types are:

Gamma rays

X-rays

Ultraviolet radiation

Visible light

Infrared radiation

Microwaves

Radio waves

Sound waves, on the contrary, do not belong to the electromagnetic spectrum, since they are another type of wave called mechanical waves (which consist of vibrations of the particles in a medium).

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Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con
Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C   and q₂ = 11.0 n C

assume the distance be r from q₁  where the electric field is zero.

distance of point from q₂  be equal to 1.8 -r

now,

        E₁ = E₂

\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}

(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}

\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume  q₁  be negative so, distance from  q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

   E₁ = E₂

\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}

(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}

\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}

1.8 =4 r

r = 0.45 m

4 0
3 years ago
Does a 2000 mercury cougar have a timing belt
Katena32 [7]
No it does not have a timing belt
5 0
2 years ago
What does the SLOPE of a Velocity-Time graph represent? *<br> 1 point
lyudmila [28]

Acceleration

Explanation:

  • In the Velocity-Time graph Velocity is taken in Y-Axis and plotted against Time in X-Axis.
  • The unit of velocity is mentioned in m/s and unit of time is mentioned in seconds.
  • Here we can analyze the instantaneous acceleration and displacement covered.
  • From the slope we can get the body accelerated.
  • From the area under the graph we can get the displacement traveled.
  • We can calculate both average and Instantaneous acceleration from the slope of this graph.

6 0
3 years ago
Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1
lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R  

Explanation:

given,

p₁ = 100 lb f/in²,               v₁ = 3.704 ft³/lb,           and T₁ = 1000 °R

p₂ = 30 lb f/in²                 n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

v₂ = 8.753 ft³/lb

work done for poly tropic process

W = \dfrac{p_1v_1-p_2v_2}{n-1}

    = \dfrac{100\times 3.704-30\times 8.753}{1.4-1}

    = 269.525 lbf/in².ft³

W = \dfrac{269.525}{5.40395} Btu/lb

   = 49.87 Btu/lb

in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

T₂ = 1291.63 °R

Final temperature is equal to 1291.63°R  

5 0
2 years ago
The motion of a free falling body is an example of __________ motion​
swat32

Answer:

accelerated

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

4 0
2 years ago
Read 2 more answers
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