C. Sonography
Sonography makes use of sound waves.
Answer:
1817.448 m
Explanation:
Given:
Initial acceleration = 2.24 m/s²
Time for acceleration = 10 s
From Newton's equation of motion
v = u + at
here v is the final velocity after the acceleration
v = 0 + 2.24 × 10 = 22.4 m/s
thus,
the displacement during acceleration
From Newton's equation of motion
where,
s is the distance
u is the initial speed = 0 m/s as starting from rest
a is the acceleration
t is the time
on substituting the respective values, we get
or
s₁ = 112 m
for case 2
time = 1.25 minute = 1.25 × 60 = 75 seconds
displacement, s₂ = v × t = 22.4 × 75 = 1680 m
for the case 3
acceleration = - 9.86 m/s²
now,
the final velocity for the second case is the initial velocity for this case
thus,
or
s₃ = 50.848 - 25.40 = 25.448 m
hence,
the total displacement = s₁ + s₂ + s₃ = 112 m + 1680 m + 25.448 m
= 1817.448 m
Answer:
8) 709.8875 J
9) The object is at 7.24375 m from the ground
10) Kinetic energy increases as the object falls.
Explanation:
We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:
h(t) = hi - (g/2) t^2
hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):
h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m
This is the position of the 10 kg object after 0.75 seconds (answer for part 9)
Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:
U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)
Part 10)
the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.
Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.