A 10-kg sled carrying a 30-kg child glides on a horizontal, frictionless surface at a speed of 6.0 m/s toward the east. The chil
d jumps off the back of the sled, propelling it forward at 20 m/s. What was the child’s velocity in the horizontal direction relative to the ground at the instant she left the sled?
1 answer:
Answer:
- 1.33m/s
Explanation:
We choose the system to be the child and the sled. The surface is friction less, which means that there are no forces exerted on the system horizontally. This means that the horizontal momentum component of the system is constant and conserved.
So we can use the conservation momentum principle to find the velocity of the child just after he leaves the sled.
This is shown in the attached file.
You might be interested in
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
#SPJ4
