Answer:
a counterclaim
Explanation:
authors purpose is what an author wrote somthing for
opinion is someones thoughts or "side" on a argument
an arguement is a battle of opinions if that makes sense
Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
Formula(2)

a =8cm=0.08m

We replace the known information in the formula (2)

I=0.8 A
Answer: The electric current in the wire is 0.8 A
Answer:
The observed wavelength on Earth from that hydrogen atom is
.
Explanation:
Given that,
The actual wavelength of the hydrogen atom, 
A hydrogen atom in a galaxy moving with a speed of, 
We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

is the observed wavelength

So, the observed wavelength on Earth from that hydrogen atom is
. Hence, this is the required solution.
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
Learn more about work done here: brainly.com/question/25573309
#SPJ1
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.