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4 years ago

Which type of solar radiation is the most powerful?

Physics

2 answers:

zaharov [31]4 years ago

5 0

Answer:

Gamma rays are the most powerful

Gamma rays are not necessarily harmful to the planet Earth, but to all the living organisms that inhabit it, as Gamma is a powerful form of radiation.
Gamma-rays are produced primarily by four different nuclear reactions: fusion, fission, alpha decay and gamma decay

GAMMA RAYS

Gamma-rays are a form of electromagnetic radiation, as are radio waves, infrared radiation, ultraviolet radiation, X-rays and microwaves. Gamma-rays can be used to treat cancer, and gamma-ray bursts are studied by astronomers.

Electromagnetic (EM) radiation is transmitted in waves or particles at different wavelengths and frequencies. This broad range of wavelengths is known as the electromagnetic spectrum. The spectrum is generally divided into seven regions in order of decreasing wavelength and increasing energy and frequency. The common designations are radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma-rays.

Gamma-rays fall in the range of the EM spectrum above soft X-rays. Gamma-rays have frequencies greater than about 1,018 cycles per second, or hertz (Hz), and wavelengths of less than 100 picometers (pm), or 4 x 10^9 inches. (A picometer is one-trillionth of a meter.)

Gamma-rays and hard X-rays overlap in the EM spectrum, which can make it hard to differentiate them. In some fields, such as astrophysics, an arbitrary line is drawn in the spectrum where rays above a certain wavelength are classified as X-rays and rays with shorter wavelengths are classified as gamma-rays. Both gamma-rays and X-rays have enough energy to cause damage to living tissue, but almost all cosmic gamma-rays are blocked by Earth's atmosphere.

cluponka [151]4 years ago

4 0

Answer: Gamma rays

Explanation:

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What is motion? give 3 examples

Tomtit [17]

Answer:

Motion definition: the action or process of moving or being moved.

Examples:

1. moving a hand

2. riding a bicycle

3. running

8 0

3 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is:

$7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}$

$p_{B,2} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is:

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}$

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0

3 years ago

A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

Where:

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In this problem we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

5 0

3 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.51 s,

Gelneren [198K]

The pot spends T = 0.185s going up and 0.185s going down past the window.
The average speed passing by the window is 2.20 m/0.185s = 11.89 m/s.
During passage, the pot increases speed by T*g = 0.185*9.81 = 1.815 m/s
The speed is therefore 12.80 m/s at the bottom of the window and 10.98 m/s at the top of the window.

The 10.98 m/s speed at the top of the window allows it to rise another 10.98^2/(2g)= 6.15 m past the top of the window

8 0

3 years ago

Why we have leap year?

Marina CMI [18]

Answer:February 29 is a date that usually occurs every four years, and is called leap day. This day is added to the calendar in leap years as a corrective measure, because the Earth does not orbit the sun in precisely 365 days.

Explanation: brainlest plz

8 0

3 years ago