They can tell what type/species the dinosaur was
A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid. Hope this helps!
Answer:
2880 N/c
Explanation:
Given that:
Charge per unit length ; λ = 4 * 10^-9
radius, r = 10
Radius, R = 0.5m
Using the relation :
2λr / 4πE0R²
Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²
Hence, we have :
2λrk/ R²
(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2
(720 ÷ 0.25)
= 2880 N/c
Givens
======
F1 = 785 N on earth
F2 = Force on Sun
F3 = Force on Star
r_earth = 6371 km
r_sun = 695 700 km
r_star = 15/2 km = 7.5 km
m_earth = 5.972 * 10^24 kg
m_sun = 2 * 10^30 kg
Formula
======
F_sun / F_earth = G *m_test * m_sun/(r_sun)^2 // G * m_test *m_earth / (re)^2
Again, there are cancellations
F_sun/F _ earth = [m_sun/m_earth ] * [(r_sun)^2 / (r_earth)]^2
Substitution and solution
===================
F_sun = 785 * (2*10^30 / 6*10^24) * (695700)^2 / (6371)^2 Note the radii do not have to be converted since the conversions would cancel anyway.
Solving this mess gives
F_sun = 3.12 * 10^12 Note I have rounded. You can put in the exact number.
This may not seem possible, but I don't think I've made an error. Now you go through the same process for the star. I leave that for you
Missing question: "<span>At what time is the velocity again zero?"
Solution:
the acceleration of the particle is
</span>
<span>the velocity of the particle is the derivative of the acceleration:
</span>
<span>where </span>
is the initial velocity, given by the problem.
So, to find when the particle's velocity is again zero, we should just put vx=0 and find t:
which has two solutions:
t=0 (beginning of motion)
and so, the particle velocity returns to zero after 20 seconds.