Answer:
I = (1.80 × 10⁻¹⁰) A
Explanation:
From Biot Savart's law, the magnetic field formula is given as
B = (μ₀I)/(2πr)
B = magnetic field = (1.0 × 10⁻¹⁵) T
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
r = 3.6 cm = 0.036 m
(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)
4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036
I = (1.80 × 10⁻¹⁰) A
Hope this Helps!!!
Answer:
The light used has a wavelenght of 4.51×10^-7 m.
Explanation:
let:
n be the order fringe
Ф be the angle that the light makes
d is the slit spacing of the grating
λ be the wavelength of the light
then, by Bragg's law:
n×λ = d×sin(Ф)
λ = d×sin(Ф)/n
λ = (3.2×10^-4 cm)×sin(25.0°)/3
= 4.51×10^-5 cm
≈ 4.51×10^-7 m
Therefore, the light used has a wavelenght of 4.51×10^-7 m.
Answer:
By the use of slow motion camera.
Explanation:
Visually, it is very hard to differentiate between an ac and dc power supply. But Since, we that In Ac supply polarity changes 100 times in a second ( because frequency of ac supply is 50 Hz generally). Whereas, Dc gives a steady power supply. So, in slow motion camera we can easily capture the flickering light tubes which won't happen in case of dc supply.
Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):
Rearranging the equation, we can write
In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is
2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have: