A nickel has a mass of 5g and a volume of approximately 0.689cm what is the approximate density of a nickel
1 answer:
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Draw a diagram to illustrate the problem as shown below.
The ball is thrown from A, rises to a maximum height at B, and returns to the original level at C.
Given:
v = 12.0 m/s, vertical component of launch velocity
u = 20 m/s, horizontal component of launch velocity.
Note that g = 9.8 m/s², the acceleration due to gravity.
Wind resistance is ignored.
Part a.
The time, t, required to reach B from A is given by
0 = 12 - 9.8t
t = 12/9.8 = 1.2245 s
Answer: The time is 1.225 s (nearest thousandth)
Part b.
The height attained is
h = 12*1.2245 - 05*9.8*1.2245²
= 7.347 m
Answer: The height attained is 7.35 m (nearest hundredth) above A.
Part c.
When the ball returns the level of point A or C at time t,
0 = 12t - 0.5*9.8t²
t(12 - 4.9t) = 0
t = 0 (point A), or
t = 4.9/2 = 2.45 s (point C)
Notice that 2.45/2 = 1.225 s, which is equal to the time to reach B from A.
Answer:
The time taken to travel from A to C is 2.45 s.
This time is twice the time taken to travel from A to B.
Part d.
The horizontal distance traveled from A to C is
d = 20*2.45 = 49 m
Answer: 49 m
Part e.
x(t) = 20t for horizontal travel
y(t) = 12t - 4.9t² for vertical travel
vy(t) = 12 - 9.8t for vertical velocity
vx(t) = 20 for horizontal velocity.
Graphs of x-t, y-t, vx-t, vy-t are shown below.
The ball is thrown from A, rises to a maximum height at B, and returns to the original level at C.
Given:
v = 12.0 m/s, vertical component of launch velocity
u = 20 m/s, horizontal component of launch velocity.
Note that g = 9.8 m/s², the acceleration due to gravity.
Wind resistance is ignored.
Part a.
The time, t, required to reach B from A is given by
0 = 12 - 9.8t
t = 12/9.8 = 1.2245 s
Answer: The time is 1.225 s (nearest thousandth)
Part b.
The height attained is
h = 12*1.2245 - 05*9.8*1.2245²
= 7.347 m
Answer: The height attained is 7.35 m (nearest hundredth) above A.
Part c.
When the ball returns the level of point A or C at time t,
0 = 12t - 0.5*9.8t²
t(12 - 4.9t) = 0
t = 0 (point A), or
t = 4.9/2 = 2.45 s (point C)
Notice that 2.45/2 = 1.225 s, which is equal to the time to reach B from A.
Answer:
The time taken to travel from A to C is 2.45 s.
This time is twice the time taken to travel from A to B.
Part d.
The horizontal distance traveled from A to C is
d = 20*2.45 = 49 m
Answer: 49 m
Part e.
x(t) = 20t for horizontal travel
y(t) = 12t - 4.9t² for vertical travel
vy(t) = 12 - 9.8t for vertical velocity
vx(t) = 20 for horizontal velocity.
Graphs of x-t, y-t, vx-t, vy-t are shown below.