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3 years ago

Gamma rays have a higher frequency than visible light waves. What can you conclude from this?

Physics

2 answers:

Finger [1]3 years ago

7 0

Answer:

Explanation:

The higher the frequency, the longer the wavelength, so if gamma rays have a higher frequency, than gamma rays have a longer wavelength than visible light waves

NeTakaya3 years ago

3 0

Answer:

i think it d

Explanation:

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Describe what it means if a chemical or physical property is periodic

vesna_86 [32]

It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.

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3 years ago

Faraday's law states that voltage can be changed by moving the coil out of the magnetic field true or false

NNADVOKAT [17]

As per Faraday's law of induction we know that induced EMF in a conducting closed loop is equal to rate of change in flux in that loop

So here we have

$V = \frac{d\phi}{dt}$

now when we move out a coil from magnetic field then in this case there will be EMF induced in that coil as here magnetic flux is changing with time linked with the coil.

Now this induced voltage will remain constant if coil is moved out uniformly

But it will not remain constant if coil is moved out with non uniform speed

So this statement is not always true

so answer must be

<u>FALSE</u>

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3 years ago

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Maurinko [17]

Answer:

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3 years ago

Read 2 more answers

Use a(t) = −9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A canyon is 800 meter

Kruka [31]

Answer:

t = 12.8 s

Explanation:

As we know that the deepest point of the canyon is 800 m

so here we will have the displacement of the rock in the canyon is same as that of depth of the canyon

So here we will say

$\Delta y = v_y t + \frac{1}{2}at^2$

$800 = 0 + \frac{1}{2}(9.8) t^2$

$800 = 4.9 t^2$

now we have

$t^2 = \frac{800}{4.9}$

$t = \sqrt{163.6}$

$t = 12.8 s$

7 0

4 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago