Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Answer:
12.32 m/s
Explanation:
Using the formula of maximum height of a projectile,
H = u²sin²Ф/2g................... Equation 1
Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity
make u the subject of the equation
u = √(2Hg/sin²Ф)............ Equation 2
Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²
Substitute into equation 2
u = √[(2×2.3×9.8)/sin²33°]
u =√ [45.08/(0.545)²]
u = 45.08/0.297
u = √(151.785)
u = 12.32 m/s
Answer:
13.2m
Explanation:
Step one:
given data
Energy= 5610J
Force F= 425N
Required
The distance traveled
Step two:
We know that work done is given as
WD= force* distance
so
5610=425*d
divide both sides by 425
d= 5610/425
d=13.2m
<span>t^2 = 1/4.9 </span>
<span>t = 0.45 sec
answer:</span><span>1 - 4.9t^2 = 0 </span>
Answer:
you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.
whatever we are, you and I will always collide.
There you go! Let me know if it helped.
:)
