A strong inductive argument must have true premises True False

A pendulum is set in motion and time. The time measured for 20 complete swings is 308.

laiz [17]

Answer:

solution

308=20 swings

?::: =1

( 308×1)÷20

308÷20

154÷10

=15.4

=15.4

6 0

3 years ago

15m/s is how many Newtons

GuDViN [60]

147.09975 newton meters per second

5 0

2 years ago

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

2 years ago

A gaseous sample of this element is bombarded by photons of various energies (in these same units). Match each photon to the res

solniwko [45]

1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B 
3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available 
4) n=1 -> infinity delta E = 11. The answer is A 
5) not absorbed would be C, as there is no transition with delta E of 8.

8 0

3 years ago

Can someone answer all of the questions?

Katyanochek1 [597]

Answer: -200

Explanation: is it math

8 0

2 years ago

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