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siniylev [52]
3 years ago
14

What is the gravitational relationship between two objects?

Physics
1 answer:
Inessa [10]3 years ago
5 0
I think the right answer would be objects pull because gravitational pull is when an object with more mass than an other object would pull the small mass object
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An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0
3 years ago
The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre
mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

       tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

   

        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

       θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

4 0
3 years ago
Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou
belka [17]
The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year. 
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather
4 0
3 years ago
I need to figure out A but i’m not sure how
Damm [24]

Answer:

cc

Explanation:

4 0
3 years ago
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
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