Answer:
a) t = 9.16*10^{-18} s
b) y = 0.402 mm
Explanation:
(a) To find the time that the particle takes to pass trough the region between parallel plates, you take into account that the horizontal component of the velocity is constant in all trajectory of the particle. Then, you use the following formula:
![t=\frac{x}{v_x}\\\\](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bx%7D%7Bv_x%7D%5C%5C%5C%5C)
x: length of the sides of the plates = 0.22m
v_x: horizontal component of the velocity = 2.4*10^6 m/s
![t=\frac{0.22m}{2.4*10^6m/s}\\\\t=9.16*10^{-8}s=91.6ns](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B0.22m%7D%7B2.4%2A10%5E6m%2Fs%7D%5C%5C%5C%5Ct%3D9.16%2A10%5E%7B-8%7Ds%3D91.6ns)
(b) To find the vertical displacement of the particle you first calculate the acceleration of the particle generated by the electric force:
![F_e=ma\\\\qE=ma\\\\a=\frac{qE}{m}\\\\a=\frac{(1.6*10^{-19}C)(33*10^3N/C)}{5.5*10^{-26}C}=9.6*10^{10}\frac{m}{s^2}](https://tex.z-dn.net/?f=F_e%3Dma%5C%5C%5C%5CqE%3Dma%5C%5C%5C%5Ca%3D%5Cfrac%7BqE%7D%7Bm%7D%5C%5C%5C%5Ca%3D%5Cfrac%7B%281.6%2A10%5E%7B-19%7DC%29%2833%2A10%5E3N%2FC%29%7D%7B5.5%2A10%5E%7B-26%7DC%7D%3D9.6%2A10%5E%7B10%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D)
where you have used that the charge is 1.6*10^-19 C (charge of an electron).
With the values of the acceleration and time you use the following kinematic equation to calculate the vertical displacement:
![y=v_{oy}t+\frac{1}{2}at^2\\\\v_{oy}=0m/s\\\\y=\frac{1}{2}(9.6*10^{10}\frac{m}{s^2})(9.16*10^{-8}s)^2=4.02*10^{-4}m=0.402mm](https://tex.z-dn.net/?f=y%3Dv_%7Boy%7Dt%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5C%5Cv_%7Boy%7D%3D0m%2Fs%5C%5C%5C%5Cy%3D%5Cfrac%7B1%7D%7B2%7D%289.6%2A10%5E%7B10%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%289.16%2A10%5E%7B-8%7Ds%29%5E2%3D4.02%2A10%5E%7B-4%7Dm%3D0.402mm)
Answer:
The incidence angle is 0.145°.
Explanation:
Given that,
Distance due to east = 119 m
Distance due to north= 47.0 km
We need to calculate the incidence angle
Using formula of incidence angle
![\tan\theta=\dfrac{x}{y}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7Bx%7D%7By%7D)
Where, y = distance due to north
x = distance due to east
Put the value into the formula
![\theta=\tan^{-1}\dfrac{119}{47\times10^3}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctan%5E%7B-1%7D%5Cdfrac%7B119%7D%7B47%5Ctimes10%5E3%7D)
![\theta=0.145^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D0.145%5E%7B%5Ccirc%7D)
Hence, The incidence angle is 0.145°.
Porque sus células están creadas diferentes que los humanos.
Answer:
![2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
29010.53917 m
Explanation:
= Density of asteroid = 2 g/cm³
V = Volume
d = Diameter = 10 km
r = Radius = ![\dfrac{d}{2}=\dfrac{10}{2}=5\ km](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7B2%7D%3D%5Cdfrac%7B10%7D%7B2%7D%3D5%5C%20km)
v = Velocity = 11 km/s
= Heat vaporization of water = ![2.26\times 10^6\ J/kg](https://tex.z-dn.net/?f=2.26%5Ctimes%2010%5E6%5C%20J%2Fkg)
= Change in temperature = 100-20
Mass is given by
![m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg](https://tex.z-dn.net/?f=m%3D%5Crho%20V%5C%5C%5CRightarrow%20m%3D%5Crho%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5C%5CRightarrow%20m%3D2000%5Cdfrac%7B4%7D%7B3%7D%5Ctimes%20%5Cpi%5Ctimes%205000%5E3%5C%5C%5CRightarrow%20m%3D1.0472%5Ctimes%2010%5E%7B15%7D%5C%20kg)
The kinetic energy is
![K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20K%3D%5Cdfrac%7B1%7D%7B2%7D1.0472%5Ctimes%2010%5E%7B15%7D%5Ctimes%2011000%5E2%5C%5C%5CRightarrow%20K%3D6.33556%5Ctimes%2010%5E%7B22%7D%5C%20J)
Heat is given by
![Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%2BmH_v%5C%5C%5CRightarrow%206.33556%5Ctimes%2010%5E%7B22%7D%3Dm%5Ctimes%20%284186%5Ctimes%20%28100-20%29%2B2.26%5Ctimes%2010%5E6%29%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B%206.33556%5Ctimes%2010%5E%7B22%7D%7D%7B4186%5Ctimes%20%28100-20%29%2B2.26%5Ctimes%2010%5E6%7D%5C%5C%5CRightarrow%20m%3D2.44156%5Ctimes%2010%5E%7B16%7D%5C%20kg)
Mass of water is ![2.44156\times 10^{16}\ kg](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B16%7D%5C%20kg)
Volume is ![\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=%5Cdfrac%7B2.44156%5Ctimes%2010%5E%7B16%7D%7D%7B10%5E3%7D%3D2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
Amount of water is ![2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
If it were a cube
![h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m](https://tex.z-dn.net/?f=h%3DV%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20h%3D%282.44156%5Ctimes%2010%5E%7B13%7D%29%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20h%3D29010.53917%5C%20m)
The height of the water would be 29010.53917 m
There is no change in the weight of either
Air has no influence on the weight of an object unless the object is accelerating down through an air column. The air resistance to the free fall of the object may have an influence on its instantaneous weight of the object.
Explanation:
The weight of an object is influenced by the mass of the object and gravity. Gravity also has influence in a vacuum. Therefore the wooden and iron piece will weight the same in a vacuum as in the air.
Learn More:
For more on weight check out;
brainly.com/question/1265220
brainly.com/question/3272477
#LearnWithBrainly