Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
![\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D-k%5Cfrac%7Bq%7D%7Br%5E2%7Dcos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bk%5Cfrac%7Bq%7D%7Br%5E2%7Dsin%5Ctheta%5C%20%5Chat%7Bj%7D%5C%5C%5C%5C%5Cvec%7BE%7D%3Dk%5Cfrac%7Bq%5E2%7D%7Br%7D%5B-cos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bsin%5Ctheta%5C%20%5Chat%7Bj%7D%5D)
(1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
The distance r is:
You replace the values of all parameters in the equation (1):
![\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B4.28%2A10%5E%7B-12%7DC%7D%7B%2810.21%2A10%5E%7B-3%7Dm%29%7D%5B-cos%2815.84%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2815.84%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7BE%7D%3D%28-3.61%5Chat%7Bi%7D%2B1.02%5Chat%7Bj%7D%29%5Cfrac%7BN%7D%7BC%7D%5C%5C%5C%5C%7C%5Cvec%7BE%7D%7C%3D%5Csqrt%7B%283.61%29%5E2%2B%281.02%29%5E2%7D%5Cfrac%7BN%7D%7BC%7D%3D3.75%5Cfrac%7BN%7D%7BC%7D)
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".
Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
• They depend solely on the load that generates it
• Two or more electrical charges interact, which can be positive or negative
• The energy source is based on the electrical voltage
The temperature of the substance giving off the heat decreases while the temperature of the substance receilving the heat increases. they leach what is called equlibrium point where heat energy can longer be exchanged hence equql temperature. this isThermal physics
