Answer:
Bnet=1.006*10^-6T
Explanation:
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
the magnetic field Bnet=
the magnetic field due this long wire is given by
B1=∨I1/
..............................1
B2=∨I2/
............................2
Bnet=
.......................3
Bnet=v/2*pi
Bnet=4*pi*10^-7/(2
)
Bnet=0.0000002*(641.72)^.5
Bnet=1.006*10^-6T
Answer:
a. 8.136 × 10¹³ km b. 2.669 × 10¹⁹ feet
Explanation:
a. What is this distance in kilometers?
Since Sirius A is 8.6 light years away from Earth and one light year = distance travelled by light in a year = 3 × 10⁸ m/ s × 365 days/year × 24 hr/day × 3600 s/hr = 9.4608 × 10¹⁵ m = 9.4608 × 10¹² km.
Then 8.6 light years = 8.6 × 1 light year
= 8.6 × 9.4608 × 10¹² km
= 81.363 × 10¹² km
= 8.1363 × 10¹³ km
≅ 8.136 × 10¹³ km
b. What is this distance in feet?
Since 1 meter = 3.28 feet,
8.6 light years = 8.1363 × 10¹² km
= 8136.3 × 10¹⁵ m
= 8136.3 × 10¹⁵ × 1 m
= 8136.3 × 10¹⁵ × 3.28 feet
= 26687.1 × 10¹⁵ feet
= 2.66871 × 10¹⁹ feet
≅ 2.669 × 10¹⁹ feet
Answer:No
Explanation:
Radiant energy is kinetic energy.
Answer:
F = 5.256 x 
Explanation:
From the work energy theorem we know that:
The net work done on a particle equals the change in the particles kinetic energy:
W = F.d, ΔK =
where:
W = work done by the force
F = Force
d = Distance travelled
m = Mass of the car
vf, vi = final and initial velocity of the car
kf, ki = final and initial kinetic energy of the car
Given the parameters;
m = 830kg
vi = 1.9 m/s
vf = 0 km/h
d = 0.285 m
Inserting the information we have:
F.d = 
F = 
F = 
F = 5.256 x 