I would say its a positive cgarge
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/
Explanation:
yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .
The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.
Answer:
v=8m/s
Explanation:
To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have
![W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2](https://tex.z-dn.net/?f=W_%7Bf%7D%3DE_%7Bk%7D%5C%5CW_%7Bf%7D%3D%5Cmu%20N%3D%5Cmu%28m_1%2Bm_1%29g%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Bm_1%2Bm_2%5Dv%5E2)
where
mu: coefficient of kinetic friction
g: gravitational acceleration
We can calculate the speed of the cars after the collision by using
Now , we can compute the speed of the second car by taking into account the conservation of the momentum
the car did not exceed the speed limit
Hope this helps!!
