A. Coming out near the South Pole and going in near the North Pole
Answer:
–77867 m/s/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = (final velocity – Initial velocity) /time
a = (v – u) / t
With the above formula, we can obtain acceleration of the ball as follow:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
a = (v – u) / t
a = (–23.9 – 34.5) / 0.00075
a = –58.4 / 0.00075
a = –77867 m/s/s
Thus, the acceleration of the ball is –77867 m/s/s.
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
a) PE=mgh=0.2*9.8*1.2=2.352 J
b) KE=PE=2.352 J
c)
m/s
Answer:
λ = 6.602 x 10^(-7) m
Explanation:
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is given as ;
y = mλD/d
Where;
D is the distance of the screen from the slits = 6.2 m
d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m
The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
y = mλD/d
So, λ = dy/mD
Thus,
λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)
λ = 6.602 x 10^(-7) m