F=ma
For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left
For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left
Therefore, the lighter (5kg) cart experiences a greater acceleration.
Complete Question:
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
Answer:
m = 0.001 M
For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/
Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>
Answer:
F = 19.375 x 10^-6 N
Explanation:
This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.
It is defined as
F = (ke * q1 * q2)/ r^2
Where,
ke = is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2
q1 = 5.0 x 10^-8 C
q2 = 1.0 x 10^-7 C
r = 5 ft = 1,524 m
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = 19.375 x 10^-6 N
