Answer:
O Distance
Explanation:
With the given information, it is possible to find the distance of the car has traveled on the home from school.
This is possible because distance is the length of path a body covers.
From the given values:
Distance = 5208 - 5200 = 8miles
It is not possible to find the displacement because no direction was given.
Also velocity is not possible since the time frame of travel was not state.
Answer:
E = 3544.44 N/C
Explanation:
Given:
- charge Q = 2.2 *10^-6 C
- Length L = 1.3 m
Find:
The Electric Field strength E @ a = 1.8 m
Solution:
- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:
dE = k*dq / r^2
- The charge Q is spread over entire length L, hence:
dq = (Q / L ) * dx
-The resulting dE:
dE = (k*Q/L)*(dx / r^2)
- point P lies on the x- axis with distance (x+a) from differential charge from:
dE = (k*Q/L)*(dx / (x+a)^2)
- Integrate dE over length 0 to L
E = (-k*Q/L)*( 1 / (x+a) )
E = (-k*Q/L)* (1 / a - 1 / (L+a))
E = (-k*Q/L)* (L / a(L+a))
E = (k*Q / a(L+a))
- Evaluate E @ a = 1.8 m
E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))
E = 3544.44 N/C
Answer:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.
Answer:1.27
Explanation:
Given
incident angle ![i=36^{\circ}](https://tex.z-dn.net/?f=i%3D36%5E%7B%5Ccirc%7D)
refracted angle ![r=27.5^{\circ}](https://tex.z-dn.net/?f=r%3D27.5%5E%7B%5Ccirc%7D)
Suppose
is the refractive index of material then using Snell's law we can write
![n_1\sin i=n_2\sin r](https://tex.z-dn.net/?f=n_1%5Csin%20i%3Dn_2%5Csin%20r)
where
=refractive index of air
![1\times \sin (36)=n_2\times \sin (27.5)](https://tex.z-dn.net/?f=1%5Ctimes%20%5Csin%20%2836%29%3Dn_2%5Ctimes%20%5Csin%20%2827.5%29)
![n_2=\dfrac{0.5877}{0.4617}](https://tex.z-dn.net/?f=n_2%3D%5Cdfrac%7B0.5877%7D%7B0.4617%7D)
![n_2=1.27](https://tex.z-dn.net/?f=n_2%3D1.27)