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
Answer: The correct answer is the weight of the wood.
Explanation:
Hope this helps
Answer:
The distance between first-order and second-order bright fringes is 12.66mm.
Explanation:
The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
(1)
Where 
is the distance between two adjacent maxima, L is the distance of the screen from the slits, 
is the wavelength and d is the separation between the slits.
The values for this particular case are:
Notice that is necessary to express L and in units of milimeters.
⇒ 
⇒ 
Finally, equation 1 can be used:
Hence, the distance between first-order and second-order bright fringes is 12.66mm.
Answer:
Fx= 50.0 Pounds : Components of the force along the x-axis
Fy= 86.6 Pounds : Component of the force along the y-axis
Explanation:
Conceptual Analysis
To find the components (Fx, Fy) of the total force (F), we apply the trigonometric concepts for a right triangle, where the perpendicular sides of the triangle are the components (Fx, Fy) of the force (F), the hypotenuse (h) is the magnitude of the total force F and β is the angle that forms the horizontal component with the hypotenuse.
Formulas
cos β : x/h : x: side adjacent to the β angle h: hypotenuse (1)
sin β = y/h : y: side opposite to the β angle h: hypotenuse (2)
Known Data
Known data
F= 1.00 * 10² pounds = 100 pounds : magnitude of total force
β = 60.0° to the x-axis. : Angle that forms the force with the x-axis
Problem Development
We apply the formula 1 to calculate horizontal component (Fx)
cos β :Fx/F
Fx= F cosβ = 100*cos 60° = 50.0 Pounds
We apply the formula 2 to calculate vertical component (Fy)
sin β = Fy/F
Fy= F sinβ = 100*sin 60° = 86.6 Pounds
Answer:
Explanation:
Given:
- mass of bullet,
- initial velocity of bullet,
- displacement of the bullet in the target,
Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.
- so, final velocity of the bullet,
Now using the equation of motion:
where:
acceleration of the bullet
<u>Now the force of resistance offered by the target in stopping it:</u>
