Answer:
Magnetic field experienced = 4.5 × 10⁻⁴ T
Explanation:
The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by
B = (μ₀I)/(2πr)
B = ?
I = 20 KA = 20000 A
r = 8.9 m
μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A
B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T
Answer:
they are constantly bouncing everywhere and creating preasure
Explanation:
Answer:
6.97 E 16
Explanation:
Frequency is a function of velocity of light to it's wavelength.
Mathematically written as
F = Velocity / wavelength
Velocity of light = 3 x 10^8
Wavelength =430 nm =430 x 10^-9 m
converting wavelength from nanaometer to meter we divide by 10^9
Frequency = (3 x 10^8)/(430 x10^-9) =6.97 E 16
Answer:
t = 2s
Explanation:
When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)
Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.
With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.
So at t = 2s, we have an instantaneous acceleration of -2m/s².
Answer:
angle is 19.85°
Explanation:
given data
grade = 36.1 %
vertically distance y = 36.1 m
horizontal distance x = 100 m
to find out
angle
solution
we know that according to question here
it make a triangle shape that contain 100 m base and height 36.1 m
and make angle with base is θ
so by triangle we know here
tanθ = y/x ..........1
so we now put x and y and find θ
tanθ = 36.1 / 100
tanθ = 0.361
θ = 19.85°
so angle is 19.85°
