Answer:
Electric Field at a distance d from one end of the wire is
Electric Field when d is much grater than length of the wire =
Explanation:
Given:
- Total charge over the length of the wire = Q
- Length of the wire = L
- Distance from one end of wire at which electric field is needed to find=d
Let dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let be the charge density of the wire
Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing we have
When d is much greater than the length of the wire then we have
1+\dfrac{L]{d}≈1
So the Magnitude of the Electric Field at point P =
gravity, it causes a spherical shape. the inner most layers are under mass amounts of pressure.
Answer:
Velocity=1.1m/s
Amplitude=0.35m
Explanation:
Given:
time 't' = 2.9s
wavelength 'λ'= 5.5m
distance 'd'=0.7m
The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.
So, T = 2 x 2.9 => 5.8 s
As we know that frequency is the reciprocal of time period, we have
f= 1/T = 1/5.8 =>0.2 Hz
In order to find how fast are the waves traveling, the velocity is given by
Velocity = f λ
V= 0.2 x 5.5 =>1.1m/s
The distance between the boat's highest point to its lowest point is double the amplitude.
Therefore , we can write
Amplitude 'A'= d/2 =>0.7/2 =>0.35m
The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
Learn more about resultant vector here: brainly.com/question/28047791
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Hello!
Use the equation for momentum:
Plug in the given mass and velocity into the equation: