Answer:
t= 1.2 hours
Explanation:
Define first di distance between the points, so
![\bar{x}_{water}=2*2=4](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D_%7Bwater%7D%3D2%2A2%3D4)
The distance is
![d= \bar{x}_{canoe}- \bar{x}_{water}](https://tex.z-dn.net/?f=d%3D%20%5Cbar%7Bx%7D_%7Bcanoe%7D-%20%5Cbar%7Bx%7D_%7Bwater%7D)
![d= 10-4 = 6miles](https://tex.z-dn.net/?f=d%3D%2010-4%20%3D%206miles)
![t = \frac{x}{v} = \frac{6}{2+3}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%7D%7Bv%7D%20%3D%20%5Cfrac%7B6%7D%7B2%2B3%7D)
![t= 1.2 hours](https://tex.z-dn.net/?f=t%3D%201.2%20hours)
Answer:
![15 m/s](https://tex.z-dn.net/?f=15%20m%2Fs)
Explanation:
For this problem we have to find the formula for acceleration.
F = ma
So since we already know force and looking for acceleration, we have to change it up a bit.
A = F / m
The force is 3000-N
A = 3000/ m
The mass is 200-kg
A = 3000/ 200
Do the division:
3000/ 200
15
A = 15
Answer:
volume is the correct answer
Explanation:
Answer:
![R_{max} = 2.125 R](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%202.125%20R)
Explanation:
If the maximum height attained by the rock is equal to the range of the rock
then we will say
![H = R](https://tex.z-dn.net/?f=H%20%3D%20R)
![\frac{v^2 sin^2\theta}{2g} = \frac{v^2 sin(2\theta)}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%20sin%5E2%5Ctheta%7D%7B2g%7D%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctheta%29%7D%7Bg%7D)
so from this we can say
![\frac{sin^2\theta}{2} = 2sin\theta cos\theta](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%5E2%5Ctheta%7D%7B2%7D%20%3D%202sin%5Ctheta%20cos%5Ctheta)
![tan\theta = 4](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%204)
![\theta = 75.96 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2075.96%20degree)
now original range is given as
![R = \frac{v^2 sin(2\theta)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctheta%29%7D%7Bg%7D)
![R = \frac{v^2 sin(2\times 75.96)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctimes%2075.96%29%7D%7Bg%7D)
![R = 0.47\frac{v^2}{g}](https://tex.z-dn.net/?f=R%20%3D%200.47%5Cfrac%7Bv%5E2%7D%7Bg%7D)
now we know that for maximum possible range we need to throw at 45 degree
![R_{max} = \frac{v^2 sin(2\times 45)}{g}](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctimes%2045%29%7D%7Bg%7D)
![R_{max} = \frac{v^2}{g}](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D)
![R_{max} = 2.125 R](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%202.125%20R)
Answer:
![\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%2015%20%5C%20m%2Fs%20%5C%20or%20%5C%2015%20%5C%20m%2As%5E%7B-1%7D%7D%7D)
Explanation:
We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.
![v_f= v_i+ at](https://tex.z-dn.net/?f=v_f%3D%20v_i%2B%20at)
In this formula,
is the final velocity,
is the initial velocity,
is the acceleration, and
is the time.
The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.
![\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s](https://tex.z-dn.net/?f=%5Cbullet%20%5C%20%20v_i%20%3D%205.0%20%5C%20m%2Fs%20%5C%5C%5Cbullet%20%5C%20%20a%3D%202%5C%20m%2Fs%5E2%5C%5C%5Cbullet%20%5C%20%20t%3D%205%20%20%5C%20s)
Substitute the values into the formula.
![v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)](https://tex.z-dn.net/?f=v_f%3D5.0%20%5C%20m%2Fs%20%2B%20%28%202%5C%20m%2Fs%5E2%20%2A%205%20%5C%20s%29)
Solve inside the parentheses.
![v_f= 5.0 \ m/s + (10 \ m/s)](https://tex.z-dn.net/?f=v_f%3D%205.0%20%5C%20m%2Fs%20%2B%20%2810%20%5C%20m%2Fs%29)
Add.
![v_f= 15 \ m/s](https://tex.z-dn.net/?f=v_f%3D%2015%20%5C%20m%2Fs)
The units can also be written as:
![v_f= 15 \ m*s^{-1}](https://tex.z-dn.net/?f=v_f%3D%2015%20%5C%20m%2As%5E%7B-1%7D)
The bicycle's final velocity is 15 meters per second.