Answer:
increase speed, decrease speed, and change direction
Explanation:
Answer:
You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.
Mathematically:
x = [83cos90+55cos141+69cos229+41cos281+61co... cm = -27 cm
y = [83sin90+55sin141+69sin229+41sin281+61si... cm = 55 cm
so d = √(x² + y²) = 61 cm
and Θ = arctan(55/-27) = -64º +180º (to get into QII) = 116º
Explanation:
Answer:
-0.10472![\frac{rad}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Brad%7D%7Bs%7D)
Explanation:
Remember that any <u><em>clockwise</em></u> rotation is considered negative, and any <u><em>counterclockwise</em></u> rotation is considered positive.
The formula for period is:
![T = \dfrac{2\pi}{\omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Comega%7D)
Solve for
and you get:
![\omega=\dfrac{2\pi}{T}](https://tex.z-dn.net/?f=%5Comega%3D%5Cdfrac%7B2%5Cpi%7D%7BT%7D)
The period is 60 seconds so make that division and you're left with -0.10472
or -0.10
if significant figures apply (negative because the rotation is clockwise).
Answer:
877
Explanation:
P1 = 150 atm
V1 = 0.12 m^3
P2 = 1.2 atm
V2 = ?
radius of each balloon, r = 0.16 m
Volume of each balloon,
![V_{2}=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
![V_{2}=\frac{4}{3}\pi 0.16^{3}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%200.16%5E%7B3%7D)
V2 = 0.0171 m^3
Let the number of balloons be N.
So,
P1 x V1 = N x P2 x V2
150 x 0.12 = N x 1.2 x 0.0171
N = 877.19
So, the number of balloons be 877.
1) Let's call
![V_S](https://tex.z-dn.net/?f=V_S)
the speed of the southbound boat, and
![V_E=V_s+3~mph](https://tex.z-dn.net/?f=V_E%3DV_s%2B3~mph)
the speed of the eastbound boat, which is 3 mph faster than the southbound boat. We can write the law of motion for the two boats:
![S_E(t)=V_E t=(V_S+3)t](https://tex.z-dn.net/?f=%20S_E%28t%29%3DV_E%20t%3D%28V_S%2B3%29t)
![S_S(t)=V_S t](https://tex.z-dn.net/?f=S_S%28t%29%3DV_S%20t)
2) After a time
![t=3~h](https://tex.z-dn.net/?f=t%3D3~h)
, the two boats are
![45~mi](https://tex.z-dn.net/?f=45~mi)
apart. Using the laws of motion written at step 1, we can write the distance the two boats covered:
![S_E(3~h)=3(V_S+3)=3V_S+9](https://tex.z-dn.net/?f=S_E%283~h%29%3D3%28V_S%2B3%29%3D3V_S%2B9)
![S_S(3~h)=3V_S](https://tex.z-dn.net/?f=S_S%283~h%29%3D3V_S)
The two boats travelled in perpendicular directions. Therefore, we can imagine the distance between them (45 mi) being the hypotenuse of a triangle, of which
![S_E](https://tex.z-dn.net/?f=S_E)
and
![S_S](https://tex.z-dn.net/?f=S_S)
are the two sides. Therefore, we can use Pythagorean theorem and write:
![45= \sqrt{(3V_S)^2+(3V_S+9)^2}](https://tex.z-dn.net/?f=45%3D%20%5Csqrt%7B%283V_S%29%5E2%2B%283V_S%2B9%29%5E2%7D%20%20%20)
Solving this, we find two solutions. Discarding the negative solution, we have
![V_S=9~mph](https://tex.z-dn.net/?f=V_S%3D9~mph)
, which is the speed of the southbound boat.