What value is bigger hecto or kilo
2 answers:
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Answer:
(a) 25 m
(b) 75 m
Explanation:
Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.
So, the speed of the jogger,
Let d be the distance covered by him in time, t s.
As distance=(speed) x (time)
So,
From equation (i)
As the jogger starts from origin, so, the distance, , also represents the position of the jogger at the time
s.
The position-time graph has been shown.
(a) From equation (ii), for t=5.0 s
So, the jogger is at a distance of 25 m from the origin.
(b) Similarly, for t=15.0 s
So, the jogger is at a distance of 75 m from the origin.
Hi there!
We can use Biot-Savart's Law for a moving particle:
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.