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3 years ago

How should the goals of a beginner differ from someone who is more advanced?

Physics

1 answer:

nevsk [136]3 years ago

6 0

Hi!

A beginner is less skilled than someone who is more advanced, and has, in most cases, less experience.

Therefore, if they each set realistic goals for their skill levels, a beginner will have less, adventurous (for lack of a better word) goals because the beginner knows they can achieve less, while an advanced person would set higher goals that are more difficult to achieve, but realistic for themselves.

Hope this helped!

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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.

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3 years ago

What would be 1/2 times 800?

IgorLugansk [536]

Answer:

The answer is 400

Explanation:

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3 years ago

An object is transported to three different planets in our solar system. Which of the following is true about that object?

Digiron [165]

Answer;

D. The object’s weight changes, but its mass stays the same.

Explanation;

Mass is the amount of matter in a object, which is measured in kilograms. Mass of an object is measured using a beam balance. It is important to note that the mass of an object or a body remains constant, and does not vary from one place to another. For instance the mass of a person on the moon will be the same as when the person is on the earth surface.
Weight on the other hand is the measurement of gravitational pull of an object. weight is measured using a spring balance and measured in Newtons. Weight varies from one place to another depending on the gravitational pull of a given surface.

8 0

3 years ago

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A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How

stepan [7]

Answer:

$6 \times 10^4 \; \rm J$ .

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ , where

$m$ is the mass of the car, and
$v$ is the speed of the car.

The $2 \times 10^3\; \rm kg$ car would have an initial KE of:

$\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J$ .

The $3 \times 10^3\; \rm kg$ car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum $p$ of an object is equal to its mass $m$ times its velocity $v$ . In other words, $p = m\cdot v$ .

Let the mass of the two cars be denoted as $m_1$ and $m_2$ , and their initial speeds $v_1$ and $v_2$ . Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as $v_3$ .

Initial momentum of the two-car system:

$\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

After the collision, both car would have a velocity of $v_3$ (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

$m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3$ .

Since momentum is conserved during the collision, the momentum of the system after the collision would also be $2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ . That is: $(m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ .

Solve for $v_3$ :

$\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}$ .

Hence, the total kinetic energy after the collision would be:

$\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}$ .

The amount of kinetic energy lost during the collision would be:

$\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}$ .

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3 years ago

Which of these would best define sound intensity? A) how fast the wave is traveling B) the frequency of the sound wave the power

faust18 [17]

Which of these would best define sound intensity?

A) how fast the wave is travelling

B) the frequency of the sound wave the power of the sound wave per unit area the mass of the wave divided by its volume

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3 years ago

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