Answer:
359 g Mn
General Formulas and Concepts:
Explanation:
<u>Step 1: Define</u>
6.53 mol Mn
<u>Step 2: Find conversion</u>
1 mol Mn = 54.94 g Mn
<u>Step 3: Dimensional Analysis</u>
<u /> = 358.758 g Mn
<u>Step 4: Simplify</u>
<em>We are given 3 sig figs.</em>
358.758 g Mn ≈ 359 g Mn
<u>Answer:</u>
a) Speed of mailbag after 3 seconds = 32.4 m/s
b) Package is 44.1 meter below helicopter
c) If the helicopter was rising steadily at 3.00 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Package is 44.1 meter below helicopter
<u>Explanation:</u>
a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Initial velocity = 3 m/s, acceleration = 9.8 and time = 3 seconds.
v = 3+9.8*3 = 32.4 m/s
Speed of mailbag after 3 seconds = 32.4 m/s
b) We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 53.1 meter.
So package is (53.1-9)meter below helicopter = 44.1 m
c) Initial velocity = -3 m/s, acceleration = 9.8 and time = 3 seconds.
v = -3+9.8*3 = 26.4 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 35.1 meter.
So package is (35.1+9)meter below helicopter = 44.1 m
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
THe compression of the ball when it strikes the floor is 5.11*10^-3m