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Simora [160]
3 years ago
13

A 1kg of compound a with specific heat capacity of 1000J is heated increasing its temp by 1 degree. How much energy has been add

ed to block
Physics
1 answer:
Orlov [11]3 years ago
3 0
This is the definition of specific heat capacity. 1000J of heat are required to raise the temperature of 1kg of the compound by 1 degree. Therefore a 1000J have been added to the block.

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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one
Harman [31]

Answer: 5.9(10)^{-8} m

Explanation:

The equation to calculate the center of mass C_{M} of a particle system is:

C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}

Where:

m_{E}=5.9(10)^{24} kg is the mass of the Earth

m_{p}=55(10)^{9} kg is the mass of 1 billion people

r_{E}=6371000 m is the radius of the Earth

r_{E-p}=6371000 m- 2m=6370998 m is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}

C_{M}=5.9(10)^{-8} m This is the displacement of Earth's center of mass from the original center.

8 0
3 years ago
Which of the following statements is TRUE about updating the exposure control plan?
iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the  reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and  necessary to reduce exposure</em>

An exposure control plan can be regarded as  the framework for compliance between the employer and the workers.

  • This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

  • This plan gives hope to workers in term of protection when working with their Employer.

  • There are some elements that is associated with  Exposure Control Plan, and theses are;
  1. Health hazards as well as  risk that is attributed to  each product in the worksite.
  2. Statement of purpose.
  3. procedures and practices in a written form
  4. Responsibilities from the Manager, CEO, designated resources and employer.

Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0
2 years ago
How far away is mars from the sun. why is is red?
Degger [83]
Mars is 139.58 million miles away from the sun. mars is red because its soil has iron oxide (rust particles) in it.
7 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
3 years ago
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