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4 years ago

A 10-solar-mass main-sequence star will produce what types of remnants?

Physics

1 answer:

Paraphin [41]4 years ago

5 0

Answer:

A neutron star.

Explanation:

A 10-solar-mass main-sequence star will produce a neutron star.

Neutron star is a celestial object having very high density, very small radius of 30 kilometers and predominantly made up of closely packed neutrons. They are formed due to gravitational collapse of a massive star when it runs out of fuel.

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A copper wire has a diameter of 4.00 x 10-2 inches and is originally 10.0 ft long. What is the greatest load that can be support

jek_recluse [69]

Complete question is;

A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.

Answer:

F_max = 28.9 lbf

Explanation:

Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.

Thus;

Elastic limit = Max stress

Formula for max stress is;

Max stress = F_max/A

Thus;

Elastic limit = F_max/A

F_max is maximum load

A is area = πr²

We have diameter; d = 4 × 10^(-2) inches = 0.04 in

Radius; r = d/2 = 0.04/2 = 0.02

Plugging in the relevant values into the elastic limit equation, we have;

2.30 × 10⁴ = F_max/(π × 0.02²)

F_max = 2.30 × 10⁴ × (π × 0.02²)

F_max = 28.9 lbf

5 0

3 years ago

If 0.5 A is flowing through a household light

stiks02 [169]

Answer:

60W

Explanation:

P=IV=0.5x120

P =60W

5 0

3 years ago

Define in own words.

MAVERICK [17]

A recurring illness. Typically lasts for a very long time and are very hard to remove.

6 0

4 years ago

Calculating the acceleration of two people pulling a box with same force and angle.

Dmitriy789 [7]

Here as we can see there we be net horizontal force acting on it

$F_x = F_1cos20 + F_2cos20$

also we know that

$F_1 = F_2 = 80 N$

now we will have

$F_x = 2(80cos20)$

$F_x = 150.35 N$

now for net force of box we know

$F_x - F_f = ma$

$150.35 - 100 = ma$

$20a = 50.35$

$a = 2.52 m/s^2$

so acceleration of the box will be 2.52 m/s/s

7 0

3 years ago

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w

FromTheMoon [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

$\lambda = \frac{1}{1523611.1112}$

$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

$\lambda = 434nm$

For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

7 0

3 years ago