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3 years ago

Why do the positions of stars change in the universe?

1 answer:

6 0

If stars never changed, then constellations wouldn't change. But the stars, including the Sun, travel in their own separate orbits through the Milky Way galaxy. The stars move along with fantastic speeds, but they are so far away that it takes a long time for their motion to be visible to us.

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Material speed of light

Cloud [144]

The question is poor. Light doesn't refract on its way THROUGH anything. It refracts at the boundary BETWEEN two different media. The effect is greatest where the ratio of the speeds of light in the two media is greatest. On your list, that would be at the boundary between air or space and glass.

3 0

2 years ago

Read 2 more answers

This is a measure of the quantity of matter.

emmainna [20.7K]

The measure of the quantity of matter would be mass. Mass is measured in kilograms. I hope this helped!:)

3 0

2 years ago

Read 2 more answers

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

$q = 1.6 \times 10^{-19} C$

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

8 0

3 years ago

Which choice correctly ranks these items from smallest to largest?

Vera_Pavlovna [14]

Answer:

your answer is: electron → carbon atom → quantum dot → E. coli bacteria cell → comma

Explanation:

6 0

3 years ago