The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity
. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

Substitute 0 m/s for
, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

The horizontal distance traveled by the bullet is given by,

Substitute 500 m/s for
and 0.5530s for t.

The bullet travels a distance of 276.5 m.
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)
Here is the correct answer of the given question above. When a person steps forward out of a small boat onto a dock, the boat recoils backward in the water and this occurs because the total momentum of the system is conserved. Hope this helps.
Answer:
529 Hz
Explanation:
595(343-20)/(343 + 20) = 529