Answer:
she must increase the current by factor of 7
Explanation:
The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,
B= μo I/(2πr)
where,
'μo' represents permeability of free space i.e 4π*10-7 N/A2
B=magnetic field
I= current
r=radius
->When r= 1cm=> 0.01m
B1 = μo
/(2π x 0.01)
->when r=7cm =>0.07m
B2 = μo
/(2π x 0.07)
Now equating both of the magnetic fields, we have
B1= B2
μo
/(2π x 0.01)= μo
/(2π x 0.07)
/
= 0.01/0.07
/
= 1/ 7
Therefore, she must increase the current by factor of 7
Answer:
= 1.9 cm
Explanation:
The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective
M = M₀ 
Where M₀ is the magnification of the objective and
is the magnification of the eyepiece.
The eyepiece is focused to the near vision point (d = 25 cm)
= 25 /
The objective is focused on the distances of the tube (L)
M₀ = -L / f₀
Substituting
M = - L/f₀ 25/
1) Let's look for the focal length of the eyepiece (faith)
= - L 25 / f₀ M
M = 400X = -400
= - 12 25 /0.40 (-400)
= 1.875 cm
Let's approximate two significant figures
= 1.9 cm
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant

let's do it for this case:

the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator

so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
Answer:
v = 3.04 m/s
Explanation:
given,
mass of the block, M = 6.6 Kg
horizontal force, F = 12.2 N
distance, L = 2.5 m
initial speed = 0 m/s
speed of the block,v = ?
we now
Work done is equal to change in Kinetic energy.
Work done = Force x displacement
W = Δ K E
Δ K E = Force x displacement


3.3 v² = 30.5
v² = 9.242
v = 3.04 m/s
speed of the block is equal to 3.04 m/s