"Asteroid" is the name we give to the huge number of small bodies
that orbit the sun, here in the inner solar system.
Their orbits are scattered all over the place. Most of them spend
most of the time between the orbits of Mars and Jupiter, but there are
many asteroids that sometimes come very close to Earth.
Answer:
The force is 
Explanation:
From the question we are told that
The length of the box is 
The width of the box is 
The height is 
The pressure experience on one of the sides is mathematically represented as
Where A is the area of the box which is mathematically evaluated as

substituting values


This pressure is equivalent to the atmospheric pressure which has a constant value of 
This implies that

=> 
=> 
Answer:
a) 
b) 
Explanation:
The frequency of the
harmonic of a vibrating string of length <em>L, </em>linear density
under a tension <em>T</em> is given by the formula:

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

Answer:
21.28 m
Explanation:
height, h = 71 m
velocity of raft, v = 5.6 m/s
let the time taken by the stone to reach to raft is t.
use second equation of motion for stone

u = 0 m/s, h = 71 m, g = 9.8 m/s^2
71 = 0 + 0.5 x 9.8 x t^2
t = 3.8 s
Horizontal distance traveled by the raft in time t
d = v x t = 5.6 x 3.8 = 21.28 m
its D
i hope this helps to future test takers. i took the test and had to retake it because their wasn't an answer