Answer:
they are constantly bouncing everywhere and creating preasure
Explanation:
Answer:
B = 5.59x10⁹ T
Explanation:
The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:
<u>We have:</u>
F = 1.4x10⁻³ N
v = 2.6x10⁶ m/s
θ = 37.0°
q = 2*p = 2*1.6x10⁻¹⁹ C
Hence, the strength of the magnetic field is:
![B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T](https://tex.z-dn.net/?f=%20B%20%3D%20%5Cfrac%7BF%7D%7Bqvsin%28%5Ctheta%29%7D%20%3D%20%5Cfrac%7B1.4%20%5Ccdot%2010%5E%7B-3%7D%7D%7B1.6%20%5Ccdot%2010%5E%7B-19%7D%20C%2A2.6%20%5Ccdot%2010%5E%7B6%7D%2Asin%2837%29%7D%20%3D%205.59%20%5Ccdot%2010%5E%7B9%7D%20T%20)
Therefore, the strength of the magnetic field is 5.59x10⁹ T.
I hope it helps you!
Answer:
Newton gives us force
F=-G*M*m/R^2
Divide by the mass of the object to get acceleration
a=F/m = GM/R^2
You are solving the distance where the force from the moon equals the force from the Earth -
a= G*(mass of earth)/R^2 - G*(mass of moon)/(384400000-R)^2=0
Or in other words
G*(mass of earth)/R^2=G*(mass of moon)/(384400000-R)^2
Divide by G*(mass of the moon) and multiply by R^2 to get terms on both sides
G*(mass of earth)/(G*(mass of moon))=R^2/(384400000-R)^2
Since the mass of the moon and earth are fixed, the left side is a number
81.3 = R^2/(384400000-R)^2 → R^2/(384400000-R)^2–81.3 = 0
simplify
R/(384400000-R) = 9.01665
R=346024000 = 346,024 km.
Explanation:
1.) D.
2.) B.
3.) B.
Hope this helps
Early death. health problems, terrible mental state(mostly physical).