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2 answers:
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Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in
PUTTING ALL VALUE IN ABOVE EQUATION
d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km
Answer:
d. equal to one-fourth the acceleration at the surface of the asteroid.
Explanation:
The explanation is attached as a picture with this answer
Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.
as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.
As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.
It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.
Answer:
b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt +Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
