Answer:
4.3 cm
Explanation:
We are given that
Width,d=70.3
Wavelength,
We have to find the width in cm of the pattern.
The angle for the first minimum m=1
The width of the pattern=
What units should be used when measuring the mass of a ladybug?
1 answer:
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a) Electric field at x = -2 m: 21,060 N/C to the left
b) Electric field at x = 2 m: 18,000 N/C to the right
c) Electric field at x = 6 m: 18,000 N/C to the left
d) Electric field at x = 10 m: 21,060 N/C to the right
e) Electric field is zero at x = 4 m
Explanation:
a)
The electric field produced by a single-point charge is given by
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
Here we have two charges of
each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.
We call the charge at x = 0 as , and the charge at x = 8 m as
.
For a point located at x = -2 m, both the fields and
produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:
b)
In this case, we are analyzing a point located at
x = 2 m
The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:
And since the sign is positive, the direction is to the right.
c)
In this case, we are considering a point located at
x = 6 m
The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:
And the negative sign indicates that the electric field in this case is towards the left.
d)
In this case, we are considering a point located at
x = 10 m
This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:
And the positive sign means the field is to the right.
e)
We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be
This means that we have to solve the equation
Re-arranging it,
So
So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)
Learn more about electric fields:
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This question is incomplete, the complete question is;
Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.
Answer:
- the magnitude of the resultant force is 557.32 kN
- the direction of resultant force is 48.29°
Explanation:
Given the data in the question and the diagram below,
First we work on the force on the left hand side.
Left Horizontal
= βgAr
here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²
we substitute
= βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N
Left Vertical
= ( βgπh² / 2 ) × W
we substitute
= [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N
Now we go to the right hand side
Right Horizontal
= βgAh
here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²
we substitute
= ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N
Right Vertical
= ( βgπh² / 4 ) × W
we substitute
= [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N
Hence
Fx = -
= 52974 N - 423792 N = 370818 N
Fy = +
= 332845.458 N + 83211.36 N = 416056.818 N
R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N
R = 557.32 kN
Therefore, the magnitude of the resultant force is 557.32 kN
Direction of resultant force;
tanθ = Fy / Fx
we substitute
tanθ = 416056.818 N / 370818 N
tanθ = 1.121997
θ = tan⁻¹( 1.121997 )
θ = 48.29°
Therefore, the direction of resultant force is 48.29°
