The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a height of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Substituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 - 4*a*6
Solving for a
a = 5.52 m/s^2
Answer:
.012
Explanation:
Take the mass of the fish and divide it by the mass of the water:
65/.30=216.667
Divide the given speed by the value we found above:
2.5/216.667=.0115
Answer can be rounded up to .012
Answer: 1. higher than it was before they started running
Explanation: As the vacationers run towards the back(stern) of the ship the exerting more pressure against the pressure exerted by the wave supporting the moving ship,the pressure exerted on the moving ship will be increased, leading to a slight increase in the speed of the ship compared to the speed before they started running towards the back(stern) of the ship.
<span>The outermost energy level of an element are called the valence shell, that holds the valence electrons. they consist of the highest energy level. In aluminum, the valence electrons are 3. </span>
Answer:
3.46 seconds
Explanation:
Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.
The equation for the centripetal force is as follows -
Where,
is the mass of the ball,
is the speed and
is the radius of the circular path which will be equal to the length of the rope.
This centripetal force will be equal to the tension in the string and thus we can write,

and, 
Thus,
m/s.
Now, the total length of circular path = circumference of the circle
Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m
Time taken to complete one revolution =
=
= 3.46 seconds.
Thus, the mass will complete one revolution in 3.46 seconds.