Answer:
B
Explanation:
Because the main focus of cognitive psychology is on the metal processes that affect behavior
1- mass
2. 3
3. Li2O
4. Of the strong attraction..................
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:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Answer:

east of south
Explanation:
Given:
- distance of the person form the initial position,

- direction of the person from the initial position,
north of east
- distance supposed to travel form the initial position,

- direction supposed to travel from the initial position, due North
<u>Now refer the schematic for visualization of situation:</u>

...............(1)

.................(2)
<u>Now the direction of the desired position with respect to south:</u>


east of south
<u>Now the distance from the current position to the desired position:</u>


