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djverab [1.8K]
3 years ago
6

Which waves in the electromagnetic spectrum are considered high energy waves ?​

Physics
2 answers:
Rom4ik [11]3 years ago
5 0
Gamma ray and x Ray I’m pretty sure. Because they have the shortest wavelengths but the highest energy wave. Hope this helped :)
shusha [124]3 years ago
3 0

Answer:

The waves with the most higher energy are Gamma rays and X-rays

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different wavelengths at which it radiates and its different intensities.

The energy of each wavelength can be determined by means of the following equation:

E = h\nu  (1)

but \nu = \frac{c}{\lambda}, therefore:

E = \frac{hc}{\lambda}  (2)

Where h is the Planck constant, c is the speed of light and \lambda is the wavelength.

Radiation is distributed along that electromagnetic spectrum according with the wavelength or frequency. Therefore, the ones with lower wavelength will have higher energy according to equation 2, since they are inversely proportional.

For example, Gamma rays, X-rays, ultraviolet rays, the visible region are the ones with the lower wavelengths and infrared, microwave and radio waves with higher wavelength.

Then, the waves with the most higher energy are Gamma rays and X-rays.

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Electric forces stay the same as distance changes electric field change true or false
Kisachek [45]

Answer:

False

Explanation:

7 0
3 years ago
A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
GalinKa [24]

Answer:

a = 52s²

Explanation:

<u>How to find acceleration</u>

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

<u>Solve</u>

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

7 0
3 years ago
The volume of an object as a function of time is calculated by v = At3+B÷t where t is time measured in seconds and v is in cubic
madam [21]

Answer:

A = m³/s³ = [L]³/[T]³ = [L³T⁻³]

B = m³s = [L³T]

Explanation:

We have the equation:

V = At³ + B/t

where, the dimensions of each variable are as follows:

V = m³ = [L]³

t = s = [T]

substituting these in equation, we get:

m³ = A(s)³ + B/s

for the homogeneity of the equation:

A(s)³ = m³

<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>

Also,

B/s = m³

<u>B = m³s = [L³T]</u>

3 0
4 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
A horizontal net force F is exerted on an object at rest. The object starts at x=0 m and has a speed of 4.0 m/s after moving 4.0
Alex_Xolod [135]

Explanation:

The solution is be found in the attachment.

8 0
3 years ago
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