Answer:
3.34kg/min²
Explanation:
Convert to the same unit of time:
120kg/h = 2kg/min
10s = 0.6min
acceleration = change in velocity / time
a = 2/0.6 = 3.34
Change unit if needed!
A. Conduct an exam. Thomas might have a drug dependence.
(Please mark Brainliest! :D)
Answer: The force needed is 140.22 Newtons.
Explanation:
The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.
The force exerted on the pellet is
F = m a
In order to calculate the acceleration, given the displacement d,
![d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%5Cimplies%20a%3D%5Cfrac%7B2d%7D%7Bt%5E2%7D)
we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:
![\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}](https://tex.z-dn.net/?f=%5Coverline%7Bv%7D%3D%5Cfrac%7B1%7D%7B2%7D%28v_%7Bend%7D-v_%7Bstart%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%28155-0%29%5Cfrac%7Bm%7D%7Bs%7D%3D77.5%5Cfrac%7Bm%7D%7Bs%7D)
This value can be used to determine the time for the pellet through the barrel:
![t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7B%5Coverline%7Bv%7D%7D%3D%5Cfrac%7B0.6m%7D%7B77.5%5Cfrac%7Bm%7D%7Bs%7D%7D%5Capprox0.00774s)
Finally, we can use the above to calculate the force:
![F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20m%5Cfrac%7B2d%7D%7Bt%5E2%7D%20%3D%200.007kg%5Ccdot%20%5Cfrac%7B2%5Ccdot%200.6%20m%7D%7B0.00774%5E2%20s%5E2%7D%5Capprox%20140.22N)
Answer:
10.2 m
Explanation:
The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:
![y=\frac{\lambda (m+\frac{1}{2})D}{d}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%5Clambda%20%28m%2B%5Cfrac%7B1%7D%7B2%7D%29D%7D%7Bd%7D)
where
y is the position of the m-th minimum
m is the order of the minimum
D is the distance of the screen from the slit
d is the width of the slit
is the wavelength of the light used
In this problem we have:
is the wavelength of the light
is the width of the slit
m = 13 is the order of the minimum
is the distance of the 13th dark fringe from the central maximum
Solving for D, we find the distance of the screen from the slit:
![D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m](https://tex.z-dn.net/?f=D%3D%5Cfrac%7Byd%7D%7B%5Clambda%28m%2B%5Cfrac%7B1%7D%7B2%7D%29%7D%3D%5Cfrac%7B%280.0857%29%280.0011%29%7D%7B%28683%5Ccdot%2010%5E%7B-9%7D%29%2813%2B%5Cfrac%7B1%7D%7B2%7D%29%7D%3D10.2%20m)