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slava [35]
4 years ago
9

How small are the wavelengths of gamma ray radiation? A. smaller than an atom B. about the size of a golf ball C. about 10 times

the size of an atom D. about the size of earth **:( thank you!!
Physics
2 answers:
Lelu [443]4 years ago
7 0
Electromagnetic waves about the size of a golf ball are used for
microwave radio communication and radar every day.

Electromagnetic waves that are about 4-1/2 inches long are used
in the microwave oven to heat leftover meatloaf and make popcorn .

Electromagnetic waves that are 10 times the size of an atom
are hard hard X-rays.

Electromagnetic waves that are the size of Earth's diameter
are used for ELF communications, including standard time
signals, and submarine communication.

Mechanical waves that are the same order of magnitude as
the Earth are seismic (earthquake) waves.

Gamma rays are electromagnetic waves with wavelengths in the
neighborhood of the same size as the atomic nucleus. To get
some idea of their frequency, just contemplate how long it would
take light ... (which can get to the moon in a couple of seconds) ...
to move from one side of an atomic nucleus to the other side.


Daniel [21]4 years ago
6 0

The correct answer is: A. Smaller than an atom.

You might be interested in
Which statement about moons is true?
kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

8 0
4 years ago
How much total energy is dissipated in 10. seconds
noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

P=\dfrac{E}{t}

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, P=I^2R

So equation (1) becomes :

E=I^2Rt

E=(0.5\ A)^2\times 4\Omega \times 10\ s

E=10\ J

So, the correct option is (3)

Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di
PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ,  the relation is

I = I₀ cos²θ

I / I₀  = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0
3 years ago
Can you help me with this paper please I will give you 20 points!
sweet [91]
1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

Hope this helps
6 0
3 years ago
Read 2 more answers
Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s
nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}

with this value of vo you calculate the max time:

t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

- - - - - - - - - - - - -- - - - - - - - - - - - - -

TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0
3 years ago
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