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AfilCa [17]
3 years ago
9

A person on shore sees light

Physics
1 answer:
AlekseyPX3 years ago
5 0

Answer:

15.7

Explanation:

Since N1 sin(angle1)= N2 sin(angle2), you can set up the equation to find the missing angle using inverse sin( 1.00 sin(21.2)/ 1.33). You would then get that the other angle is equal to 15.7 or 15.8 if you round up. I hope this helped! :)

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A sample of Bismuth-212 has a mass of 2.64 grams (g) after 121 seconds (s). What was the initial mass of the sample if Bismuth-2
uysha [10]
The mass of a radioactive element at time t is given by
m(t) = m_0 ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }
where m_0 is the mass at time zero, while t_{1/2} is the half-life of the element.

In our problem, m(t)=2.64 g, t=121.0 s and t_{1/2}=60.5 s, so we can find the initial mass m_0:
m_0= \frac{m(t)}{ (\frac{1}{2})^{t/t_{1/2}} } = \frac{2.64 g}{( \frac{1}{2} )^{121/60.5}} =4 \cdot 2.64 g=10.56 g
4 0
3 years ago
What force is required to accelerate a body with the mass of 15 kilograms at a rate of 8m/s2?
Tom [10]
Force = mass * acceleration
F = 15 *8 = 120 Newton
8 0
3 years ago
Which of the following are correct statements about the way an atom is put
Ann [662]

Answer:

valenc e shell

Explanation:

4 0
3 years ago
When heat is added to an object, what is happening to the item at the atomic level? (Check all that apply)
3241004551 [841]

Answer:

a

Explanation:

heat is energy, energy cannot be made or destroyed but transferred

6 0
3 years ago
Read 2 more answers
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
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