Answer:
The time taken by the brick to hit the ground, t = 0.84 s
Explanation:
Given that,
A brick falls from a height, h = 3.42 m
The initial velocity of the brick is zero.
Since the brick is under free-falling. The time equation of a free-falling body when the displacement is given is
t = 
where,
h - height from surface in meters
g - acceleration due to gravity
on substituting the values in the above equation,
t = 
= 0.84 s
Hence, time taken by the brick to hit the ground is t = 0.84 s
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.
Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.
So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
.
etc (: I hope this helps.
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
Distance will be 49.34 m
Explanation:
We have given wavelength 
Diameter of the antenna d = 2.7 m
Range L = 7.8 km = 7800 m
We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D
We know that distance is given by 
So distance D will be 49.34 m
