Answer:T=116.84 N
Explanation:
Given
Weight of hiker =1040 N
acceleration
Force exerted by Rope is equal to Tension in the rope
The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to
1 answer:
Answer:
B. i=2.79A
C. F=0.066N
Explanation:
A) By the right hand rule we have that
F=iL x B
F=iLBsin(α)
If the wire jump toward the observer the top pole face is the magnetic southpole.
B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have
C) Now the length of the wire that feels B is
and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)
I hope this is useful for you
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Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C
(a) Both the girl and the boy have the same nonzero angular displacement.
Explanation:
The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by
where
is the angular speed
t is the time interval
For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of is the same for the boy and the girl.
Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.
(b) The girl has greater linear speed.
Explanation:
The linear (tangential) speed of a point along the merry-go-round is given by
where
is the angular speed
r is the distance of the point from the centre of the merry-go-round
In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.