Explanation:
∆x=300 m×2
∆t=1.5 s
v=∆x/∆t → v=2×300/1.5 = 400 m/s
Answer:
If you mean Lewis dot diagrams, aka electron-dot diagrams, then these are diagrams that show the bonding between atoms of a molecule, and the lone pairs of electrons that may exist in the molecule.
Explanation:
Answer:
The magnetic field along x axis is
![B_{x}=1.670\times10^{-10}\ T](https://tex.z-dn.net/?f=B_%7Bx%7D%3D1.670%5Ctimes10%5E%7B-10%7D%5C%20T)
The magnetic field along y axis is zero.
The magnetic field along z axis is
![B_{z}=3.484\times10^{-10}\ T](https://tex.z-dn.net/?f=B_%7Bz%7D%3D3.484%5Ctimes10%5E%7B-10%7D%5C%20T)
Explanation:
Given that,
Length of the current element ![dl=(0.5\times10^{-3})j](https://tex.z-dn.net/?f=dl%3D%280.5%5Ctimes10%5E%7B-3%7D%29j)
Current in y direction = 5.40 A
Point P located at ![\vec{r}=(-0.730)i+(0.390)k](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%3D%28-0.730%29i%2B%280.390%29k)
The distance is
![|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}](https://tex.z-dn.net/?f=%7C%5Cvec%7Br%7D%7C%3D%5Csqrt%7B%280.730%29%5E2%2B%280.390%29%5E2%7D)
![|\vec{r}|=0.827\ m](https://tex.z-dn.net/?f=%7C%5Cvec%7Br%7D%7C%3D0.827%5C%20m)
We need to calculate the magnetic field
Using Biot-savart law
![B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_%7B0%7D%7D%7B4%5Cpi%7D%5CtimesI%5Ctimes%5Cdfrac%7B%5Cvec%7Bdl%7D%5Ctimes%5Cvec%7Br%7D%7D%7B%7C%5Cvec%7Br%7D%7C%5E3%7D)
Put the value into the formula
![B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}](https://tex.z-dn.net/?f=B%3D10%5E%7B-7%7D%5Ctimes5.40%5Ctimes%5Cdfrac%7B%280.5%5Ctimes10%5E%7B-3%7D%29%5Ctimes%28-0.730%29i%2B%280.390%29k%7D%7B%280.827%29%5E3%7D)
We need to calculate the value of ![\vec{dl}\times\vec{r}](https://tex.z-dn.net/?f=%5Cvec%7Bdl%7D%5Ctimes%5Cvec%7Br%7D)
![\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k](https://tex.z-dn.net/?f=%5Cvec%7Bdl%7D%5Ctimes%5Cvec%7Br%7D%3D%280.5%5Ctimes10%5E%7B-3%7D%29%5Ctimes%28-0.730%29i%2B%280.390%29k)
![\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})](https://tex.z-dn.net/?f=%5Cvec%7Bdl%7D%5Ctimes%5Cvec%7Br%7D%3Di%280.350%5Ctimes0.5%5Ctimes10%5E%7B-3%7D-0%29%2Bk%280%2B0.730%5Ctimes0.5%5Ctimes10%5E%7B-3%7D%29)
![\vec{dl}\times\vec{r}=0.000175i+0.000365k](https://tex.z-dn.net/?f=%5Cvec%7Bdl%7D%5Ctimes%5Cvec%7Br%7D%3D0.000175i%2B0.000365k)
Put the value into the formula of magnetic field
![B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}](https://tex.z-dn.net/?f=B%3D10%5E%7B-7%7D%5Ctimes5.40%5Ctimes%5Cdfrac%7B%280.000175i%2B0.000365k%29%7D%7B%280.827%29%5E3%7D)
![B=1.670\times10^{-10}i+3.484\times10^{-10}k](https://tex.z-dn.net/?f=B%3D1.670%5Ctimes10%5E%7B-10%7Di%2B3.484%5Ctimes10%5E%7B-10%7Dk)
Hence, The magnetic field along x axis is
![B_{x}=1.670\times10^{-10}\ T](https://tex.z-dn.net/?f=B_%7Bx%7D%3D1.670%5Ctimes10%5E%7B-10%7D%5C%20T)
The magnetic field along y axis is zero.
The magnetic field along z axis is
![B_{z}=3.484\times10^{-10}\ T](https://tex.z-dn.net/?f=B_%7Bz%7D%3D3.484%5Ctimes10%5E%7B-10%7D%5C%20T)
To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
![y = \frac{1}{2} at^2+v_0t+y_0](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2%2Bv_0t%2By_0)
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
![v = \frac{x}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
x = Displacement
t = time
With the data we have we can find the time as well
![v = \frac{x}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
![t = \frac{x}{v}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%7D%7Bv%7D)
![t = \frac{18.6}{34}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B18.6%7D%7B34%7D)
![t = 0.547s](https://tex.z-dn.net/?f=t%20%3D%200.547s)
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
![y = \frac{1}{2} at^2+v_0t+y_0](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2%2Bv_0t%2By_0)
![y = \frac{1}{2} gt^2+0+0](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%2B0%2B0)
![y = \frac{1}{2} gt^2](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
![y = \frac{1}{2} 9.8*0.547^2](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%209.8%2A0.547%5E2)
![y = 1.46m](https://tex.z-dn.net/?f=y%20%3D%201.46m)
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
Answer:
Air at higher altitude is under less pressure than air at lower altitude because there is less weight of air above it, so it expands (and cools), while air at lower altitude is under more pressure so it contracts (and heats up).
Explanation:
Hope that helped