Answer:
<h3>The answer is 8.91 m/s²</h3>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
![a = \frac{f}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bf%7D%7Bm%7D%20%20%5C%5C%20)
f is the force
m is the mass
From the question we have
![a = \frac{343}{38.5} = \frac{98}{11} \\ = 8.909090...](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B343%7D%7B38.5%7D%20%20%3D%20%20%5Cfrac%7B98%7D%7B11%7D%20%20%5C%5C%20%20%3D%208.909090...)
We have the final answer as
<h3>8.91 m/s²</h3>
Hope this helps you
That's efficiency. There's no law that it must be stated in percent.
Answer:
C) Check that the numerical answer is reasonable and the units are what you expected.
Explanation:
4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector
![\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cvec%20r_0%2B%5Cvec%20v_0t%2B%5Cdfrac12%5Cvec%20at%5E2)
where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with
![r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t](https://tex.z-dn.net/?f=r_x%3D%5Cleft%283.5%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t)
![r_y=15.5\,\mathrm m-\dfrac g2t^2](https://tex.z-dn.net/?f=r_y%3D15.5%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2)
The Coyote hits the ground when
:
![15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s](https://tex.z-dn.net/?f=15.5%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2%3D0%5Cimplies%20t%3D1.8%5C%2C%5Cmathrm%20s)
4b. Here we evaluate
at the time found in (4a).
![r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m](https://tex.z-dn.net/?f=r_x%3D%5Cleft%283.5%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%281.8%5C%2C%5Cmathrm%20s%29%3D6.3%5C%2C%5Cmathrm%20m)
5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is
![r_y=1.52\,\mathrm m-\dfrac g2t^2](https://tex.z-dn.net/?f=r_y%3D1.52%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2)
We find the shell hits the ground at
![1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s](https://tex.z-dn.net/?f=1.52%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2%3D0%5Cimplies%20t%3D0.56%5C%2C%5Cmathrm%20s)
5b. The horizontal component of the bullet's position vector is
![r_x=v_0t](https://tex.z-dn.net/?f=r_x%3Dv_0t)
where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:
![3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}](https://tex.z-dn.net/?f=3500%5C%2C%5Cmathrm%20m%3Dv_0%280.56%5C%2C%5Cmathrm%20s%29%5Cimplies%20v_0%3D6300%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D)
8500 Hz and Longitudinal
Speed = frequency x wavelength
Speed of sound at 20 degrees Celsius is approximately 340 m/s