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Masja [62]
3 years ago
13

What are alkaline earth metals used for?

Physics
2 answers:
Tomtit [17]3 years ago
6 0

Answer:

C firework

Explanation:

from Quizlet

Makovka662 [10]3 years ago
4 0
The answer is C. Fireworks
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How do you know an object is in motion?
puteri [66]
C
because an object need to be in motion to make any kind of movement
4 0
3 years ago
A car moving at 16.0 m/s, passes an observer while its horn is pressed. the velocity of sound is 343 m/s and the frequency of th
Arada [10]
We are asked to solve for the frequency heard when a car is coming towards the observer. The car is moving at 16 m/s and the velocity of the sound is 343 m/s where the car horns at 583 Hz. We will use Doppler's Effect formula in calculating the unknown frequency such that the solution is shown below:
Fl = (V + Vl) * Fs / (V - Vs)
FL = (343 + 0)*583 / (343 - 16)
FL = 611. 53 Hertz

The answer for the frequency of the observer is 611.53 hertz.
5 0
3 years ago
A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0
viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
KE = \frac{1}{2}mv^2 \\\\PE = mgh

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

E_f = \frac{1}{2}mv_f^2

And:
W_A = E_i - E_f

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) -  \frac{1}{2}mv_f^2

Solving for the work done by air resistance:
W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) -  \frac{1}{2}(.450)(19.89^2)

W_A = \boxed{42.552 J}

8 0
2 years ago
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Svetlanka [38]

<u>Answer:</u>

 Mass of base ball  m_b=3.992*10^{14}kg  

<u>Explanation:</u>

  Circumference of baseball = 2πr = 23 cm

  So radius of baseball = 3.66 cm = 3.66*10^{-2} m

   Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

   Mass of proton = 10^{-27} kg  

   Diameter of proton = 10^{-15} m

   Radius of proton =  5*10^{-16} m

   Volume of ball = \frac{4}{3} \pi r^3

   Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.    

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       So mass of base ball  m_b=3.992*10^{14}kg              

5 0
4 years ago
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Choose the statements that are true about bond strength.
vodka [1.7K]
I believe that the answer is a.
7 0
3 years ago
Read 2 more answers
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