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3 years ago

A middle-age star that burns hydrogen is a A.neutron star B.white dwarf c.main sequence star D.pusar

1 answer:

4 0

Hello!

The correct answer is C.

As we should know, a neutron star is a small cluster of neutrons. It is neither middle aged or able to burn hydrogen.
A white dwarf is a “mock planet” and has no impact on hydrogen.
A pulsar is a form of a neutron star, making C the only possible answer.
Main sequence stars are both middle aged and able to use hydrogen to form other substances.

I hope I helped!

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Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?

alexdok [17]

The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.

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3 years ago

What does the atomic number of an atom tell us?

blondinia [14]

Answer:

The number of protons in the atom!

Explanation:

8 0

3 years ago

Read 2 more answers

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

3 years ago

An astronaut, of total mass 86.0 kg including her suit, stands on a spherical satellite of mass 360 kg, both at rest relative a

Afina-wow [57]

Answer: 0.56 m/s

Explanation:

Hi, to answer this question we have to apply the formula of the conservation of momentum.

m1 v1 = m2 v2 (because the system is stationary at the beginning)

Where:

m1 = mass of the astronaut

v1= velocity of the astronaut

m2= mass of the satellite

v2= velocity of the satellite

Replacing with the values given and solving:

86 kg (2.35m/s) = 360 kg v2

202.1 kgm/s=360kg v2

202.1kgm/s /360kg =v2

v2 = 0.56 m/s

Feel free to ask for more if needed or if you did not understand something.

5 0

3 years ago