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3 years ago

What are the characteristics of high energy waves?

Physics

2 answers:

GalinKa [24]3 years ago

5 0

The answer is D! :) hope it helps

sladkih [1.3K]3 years ago

4 0

The relationship between wavelength, frequency and energy of Electromagnetic Radiation is given by E = hf = hc/lamba -------(1) So from (1) there's a linear relationship between E and f. The higher the frequency, f, the higher the energy E. Also from (1) it is obvious that the lower the wavelength, lambda, the higher the energy, E. This means the answer is D.

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If an object looses electrons, then have been transferred

slava [35]

If a neutral object loses some electrons, then it will possess more protons

4 0

3 years ago

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

$18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2$

$t_3 = 36.213s$

The time when you have passed it would be:

$20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2$

$t_4 = 38.172$

The difference between the two times would be:

$t_4-t_3 = 38.172-36.213 \approx 2s$

Therefore the correct answer is C.

4 0

3 years ago

Dfine ' weight of moist air '

grandymaker [24]

Answer:

First of all, “moist air” is air with a high water vapor content. Water vapor, the invisible, gaseous form of water, occurs in highly variable amounts in the atmosphere. Water is composed of a hydrogen atom and two oxygen atoms (H2O) and has a molecular weight of 18 grams per mole.

6 0

2 years ago

An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.50 mT. The angular momentu

liberstina [14]

Answer:

Explanation:

The angular momentum of electron mvR = 6 x 10⁻²⁵ Js

Magnetic field B = 2.5 x 10⁻³ T

radius of circular path R = mv / Bq

where m is mass , v is velocity and q is charge on electron

R² = mvR / Bq

R² = 6 x 10⁻²⁵ / 2.5 x 10⁻³ x 1.6 x 10⁻¹⁹

= 1.5 x 10⁻³

R = 3.87 x 10⁻² m

mvR = 6 x 10⁻²⁵

v = 6 x 10⁻²⁵ / mR

= 6 x 10⁻²⁵ / 9.1 x 10⁻³¹ x 3.87 x 10⁻²

= .17 x 10⁸

= 17 x 10⁶ m/s

6 0

3 years ago

Talia is on a road trip with some friends. In the first 2 hours, they travel 100 miles. Then they hit traffic and go only 30 mil

emmainna [20.7K]

Since Speed, V = Distance/Time
Average speed = Total Distance/Total Time

From the given data, Total Distance = 100 + 30 + 75 miles
and Total Time = 2 + 1 + 1 hours

Average Speed = 205/4
Average Speed = 51.25 mph ( or 51mph to the nearest whole number)

3 0

3 years ago

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