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3 years ago

The picture below shows the position of Earth and two stars.

Physics

2 answers:

alexgriva [62]3 years ago

8 0

The answer is:

18 years.

The explanation:

when Star 1 is 34 light years from Earth and Star 2 is 52 light years from Earth. that means the distance between star 1 and star 2 is 52-34 = 18 years so, If Star 2 explodes as a supernova so, the explosion would take 18 years to be seen from a planet orbiting Star 1.

abruzzese [7]3 years ago

3 0

The picture below shows the position of Earth and two stars.Star 1 is 34 light years from Earth and Star 2 is 52 light years from Earth. If Star 2 explodes as a supernova, I believe that A. 18 light years would pass before the explosion is seen from a planet orbiting Star 1.
But I am not sure, sorry. :/

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In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

barxatty [35]

Answer:

The mass of blood is $m= 5.7876kg$

The number of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the question we are told that

The volume of blood is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of the blood is $\rho_b = 1060 kg/m^3$

% of blood that is cell is = 45.0%

% of the blood that is plasma is = 55.0%

density of blood cell is $\rho_d = 1125kg/m^3$

% of cell that are white is = 1%

% of cell that is red is = 99%

The diameter of the red blood cell is = $7.5 \mu m = 7.5*10^{-6}m$

The radius of the red blood cell is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

Generally the mass is mathematically represented as

$m = \rho_b * V_b$

Substituting value

$m = 1060 * 0.00546$

$m= 5.7876kg$

Mass of cell is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The volume of white blood cell is $V_w$ = 1% of volume of cells

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The volume of red blood cells is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The number of red blood cell is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The Number of white blood cell is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The total number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

4 years ago

Read 2 more answers

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex

Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, $d = 1.8 \ m$