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3 years ago

Which part of the electromagnetic spectrum has a higher frequency than ultraviolet light?

Physics

2 answers:

almond37 [142]3 years ago

7 0

Shortwave radios have a higher frequency than ultraviolet light

kati45 [8]3 years ago

5 0

The correct answer here would be C or the third option Gamma rays

hope this helps.

Plz mark brainliest :)

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Write physical quantities and its unit

miv72 [106K]

length= metre

mass= kg

time= second

temperature = kelvin

current= ampere

luminous intensity= candela

Amount of substance = mole

etc

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7 0

3 years ago

WILL MARK BRAINLIEST PLS HELP

KiRa [710]

Answer:

Our planet's rotation produces a force on all bodies moving relative to theEarth. Due to Earth's approximately spherical shape, this force is greatest at the poles and least at the Equator. The force, called the "Coriolis effect," causes the direction of winds and ocean currents to be deflected.

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4 0

3 years ago

Which best explains why water boils in a pot sitting over fire?

gregori [183]

The Thermal energy from the fire moves to the water in the forms of heat.

Thermal energy is the energy that comes from heat which generated by the movements of tiny particles of an obejct. The faster the particles move, the more heat energy it will generates

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7 0

3 years ago

Read 2 more answers

An electron is trapped in a one-dimensional infinite well of width 340 pm and is in its ground state. What are the (a) longest,

Nesterboy [21]

Answer:

(a) 1.2703×10⁻⁷ m

(b) 4.7636×10⁻⁸ m

Explanation:

Given:

Width of the infinite well, L = 340 pm = 340×10⁻¹² m.

The formula for energy of the electron in nth state is:

$E_n=\frac {n^2\times h^2}{8mL^2}$

The expression for the difference in energy between the levels having quantum numbers n(initial) to n(final) is:

$\Delta E_n=\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}$

According to Planks theory:

E = hv

where, v is the frequency

Also,

Frequency×Wavelength = Speed of light

So,

$E=\frac {hc}{\lambda}$

$\lambda=\frac {hc}{E}$

Also, using energy from above formula as:

$\lambda=\frac {hc}{\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}}$

$\lambda=\frac {c\times {8mL^2}} {({n_f}^2-{n_i}^2)\times h}}$

For longest wavelength ni = 1 and nf = 2

m= mass of the electron = 9.1 ×10⁻³¹kg

c = 3×10⁸m/s

h = 6.625×10⁻³⁴J.sec

$\lambda_{Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({2}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Longest wavelength = 1.2703×10⁻⁷ m

For second longest wavelength ni = 1 and nf = 3

$\lambda_{Second\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({3}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Second longest wavelength = 4.7636×10⁻⁸ m

For third longest wavelength ni = 1 and nf = 4

$\lambda_{Third\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({4}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Third longest wavelength = 2.5406×10⁻⁸ m

3 0

3 years ago

A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti

Kitty [74]

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

The law of conservation of energy:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

The law of conservation of energy:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$

$\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}$

8 0

3 years ago