Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level = 
Now
Intensity level ( or Loudness)is




Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels
You haven't told us what the passing percentage is on the exam,
or what the passing percentage is for the semester, or any of that.
Answer:
a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) The total distance traveled by the baseball was 108.7 m.
Explanation:
a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

Where:
: is the final height =?
: is the initial height = 1 m
: is the initial vertical velocity = v₀sin(45)
v₀: is the initial velocity = 32.5 m/s
g: is the gravity = 9.81 m/s²
t: is the time
First, we need to find the time by using the following equation:

Now, the height is:
Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) To find the distance traveled by the baseball first we need to find the time of flight:



By solving the above quadratic equation we have:
t = 4.73 s
Finally, with that time we can find the distance traveled by the baseball:

Hence, the total distance traveled by the baseball was 108.7 m.
I hope it helps you!
It is the same if a white father and a white mother give birth to a black kid, so the mother accuses the father for having (let's say mating) with a black women.
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s