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3 years ago

What is dark matter consisted of?

Physics

2 answers:

Arturiano [62]3 years ago

6 0

Answer:

Dark matter is composed of particles that do not absorb, reflect, or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material that cannot be seen directly. We know that dark matter exists because of the effect it has on objects that we can observe directly.

Explanation:

viva [34]3 years ago

3 0

Answer:

Explanation:

You might be interested in

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

3 years ago

Which two objects have stored energy?

Furkat [3]

3) a stretched rubber band

6 0

4 years ago

Read 2 more answers

The two conducting rails in the drawing are tilted upwards so they each make an angle of 30.0° with respect to the ground. The v

dolphi86 [110]

Answer:

The current flows through the rod is 14.9 A.

Explanation:

Given that,

Magnetic field = 0.045 T

Mass of aluminum rod = 0.19 kg

Length = 1.6 m

Angle = 30.0°

We need to calculate the force

Using resolving force

$F\cos\theta=mg\sin\theta$

$F=mg\tan\theta$

Put the value into the formula

$F=0.19\times9.8\times\tan30$

$F=1.075\ N$

We need to calculate the current flows through the rod

Using formula of magnetic force

$F=iLB$

$i=\dfrac{F}{LB}$

Put the value into the formula

$i=\dfrac{1.075}{1.6\times0.045}$

$i=14.9\ A$

Hence, The current flows through the rod is 14.9 A.

8 0

3 years ago

Can someone please help me answer 14 and 15??

ziro4ka [17]

14. The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth's atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.

15.
About 5.4 billion years from now, the sun will have exhausted all of its hydrogen. The sun's core will get really hot and dense, thus shrinking; however, the outer region of the sun will expand and grow. ... Even if the expanding dying sun doesn't reach Earth, the sun's high temperatures will completely burn the planet.

7 0

4 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

6 0

4 years ago