Answer:
ANSWER BELOW I
I
V
Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in
m/s2
. For example, for Earth, 445 N = 45.4 × 9.8
m/s2
:Notice that the x-axis values will be gravity in
m/s2
, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).
Answer:
5 minutes ago A bird flies 6km east, then 4km west. Find the total distance traveled and displacement (d) after this flight. Find the total distance traveled and displacement (d) after this flight.
Explanation:
Answer:
A. 0.025inches= 1m
So 75in= 0.025*75= 1.88m
B 3.45*10^6yrs= 3.45E6* 365x 86900s
= 1.09*10^8s
C. 62ft/day= 62* 0.3048/86900= 2.1*10^-4
D. 2.2*10^4mi² x ( 1609.3)²
= 6.1*10^7m²
Answer:
After 3 hours.
Explanation:
This is a encounter problem.
We start by writing the position equation of both objects.
We assume that acceleration is equal to zero in both boats ⇒ They move with constant speed
We put the origin of coordinates in the harbour so after 6 hours the first boat travels ![7\frac{mi}{h}.6h=42mi](https://tex.z-dn.net/?f=7%5Cfrac%7Bmi%7D%7Bh%7D.6h%3D42mi)
The first boat will be 42 mi far from the harbour.
The position equation for a motion with acceleration equal to zero is :
![x(t)=x_{0}+v(t-t_{i})](https://tex.z-dn.net/?f=x%28t%29%3Dx_%7B0%7D%2Bv%28t-t_%7Bi%7D%29)
Where
is initial position
v is the speed
t is time and
is initial time
Then
is time variation
For the first boat :
![x1(t)=42mi+7\frac{mi}{h}.t](https://tex.z-dn.net/?f=x1%28t%29%3D42mi%2B7%5Cfrac%7Bmi%7D%7Bh%7D.t)
Because we set ![t_{i}=0 h](https://tex.z-dn.net/?f=t_%7Bi%7D%3D0%20h)
For the second one :
![x2(t)=21\frac{mi}{h}.t](https://tex.z-dn.net/?f=x2%28t%29%3D21%5Cfrac%7Bmi%7D%7Bh%7D.t)
In the encounter : x1(t) = x2(t) ⇒
![42mi+7\frac{mi}{h}.t=21\frac{mi}{h}.t](https://tex.z-dn.net/?f=42mi%2B7%5Cfrac%7Bmi%7D%7Bh%7D.t%3D21%5Cfrac%7Bmi%7D%7Bh%7D.t)
![14\frac{mi}{h}.t=42mi\\ t=\frac{42}{14}.h\\ t=3h](https://tex.z-dn.net/?f=14%5Cfrac%7Bmi%7D%7Bh%7D.t%3D42mi%5C%5C%20t%3D%5Cfrac%7B42%7D%7B14%7D.h%5C%5C%20t%3D3h)
So after 3 hours they have the same position relative to the harbour
We can check by replacing the value t = 3h in both position equations
For x1(t) :
![x1(3h)=42mi+7\frac{mi}{h}.(3h) \\x1(3h)=63mi](https://tex.z-dn.net/?f=x1%283h%29%3D42mi%2B7%5Cfrac%7Bmi%7D%7Bh%7D.%283h%29%20%5C%5Cx1%283h%29%3D63mi)
For x2(t) :
![x2(3h)=21\frac{mi}{h}.(3h)\\x2(3h)=63mi](https://tex.z-dn.net/?f=x2%283h%29%3D21%5Cfrac%7Bmi%7D%7Bh%7D.%283h%29%5C%5Cx2%283h%29%3D63mi)
formula for wavelength = speed/frequency
So 1500/200 = 7.5 meters