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2 years ago

What is gamma radiation composed of?

Physics

2 answers:

Montano1993 [528]2 years ago

7 0

Answer:

Explanation:

$\lim_{n \to \infty} a_n \sqrt[n]{x}$ U 238+plotonum decay

Paha777 [63]2 years ago

3 0

Gamma radiation is composed of the stable form of nuclei and most important Helium Atom is released in this reaction !

gamma radiation is due to fission in which the radioactive substance is bombarded and as a result a stable form of that element is formed and a Helium atom !

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Rogue waves are created along the "Wild Coast" off the southeast coast of ________, where the Agulhas Current flows directly aga

Vedmedyk [2.9K]

Answer:

Africa

Explanation:

A rogue wave refers to the wave that is twice the height of a significant wave occurring in a particular area. The significant wave height is generally referred to as the mean of the largest one-third of waves existing at a particular time period. In simple words, a rogue wave is much larger than any other waves that occur at the proximity of the same time.

This rough wave describes the interaction between the ocean and sea current and swelling of waves. It takes place when the large swells in the ocean, also known as the Antarctic storms, strikes with the rapidly traveling Agulhas current, and the curved water current focuses on the energy of the waves.

Thus, these Rogue waves are often generated along the southeastern coastal regions of Africa, where there occurs the convergence of Antarctic storm waves and Agulhas Current.

4 0

2 years ago

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

Law Incorporation [45]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is $r=6940 km=6.94 \cdot 10^6 m$ , the gravitational force is $F=9.21 \cdot 10^4 N$ and the mass of the Earth is $m_1=5.98 \cdot 10^{24} kg$ , therefore we can rearrange the previous equation to find m2, the mass of the telescope:
$m_2 = \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg$

6 0

3 years ago

Read 2 more answers

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

Lena [83]

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$ , $2$\hbar$ , $3$\hbar$$ , etc. Bosons particles always have asymmetric wave function and there is exchange of particles.

1) It does not obey Fermi_ Dirac statistics

2) It obeys Bose-Einstein statistics

3) The object can have intrinsic spin $2$\hbar$

4) Yes the Bosons particle is always symmetric with exchange of particles

5) No Bosons particle are symmetric and not asymmetric

5 0

2 years ago

Choose all the answers that apply. Oxygen is _____. transported by blood picked up in the alveoli released into the air by the l

aleksklad [387]

ANSWER:

- Transported by blood picked up in the alveoli
- Pumped to cells by ventricles

Hope this helps! :)

4 0

2 years ago

Read 2 more answers

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$