A charged particle moving through a magnetic field at right angles to the field with a speed of 24.7 m/s experiences a magnetic
1 answer:
Answer:
Explanation:
let the charge is q. velocity, v = 24.7 m/s
magnetic force, F = 2.38 x 10^-4 N
Let the magnetic field is B.
Velocity, v' = 5.64 m/s
angle, θ = 21.2°
The force experienced by a charged particle placed in a magnetic field is given by
F = q x v x B x Sinθ
in first case
2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)
in second case
F = q x 5.64 x B x Sin 21.2° .... (2)
Divide equation (2) by equation (1), we get
F = 1.97 x 10^-5 N
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Answer:
v = 12 m/s
Long, boring, and convoluted explanation:
First, let's lay out our information.
- <em>max height = 3.7 m</em>
- <em>0 = 45°</em>
<em>- gravitational acceleration constant = 9.8 </em><em />
<em />
Since the puma leaves the ground at a <em>
45
°</em> angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is <em>3.7m</em> because the puma can reach up to that height. Gravity is always given to be <em>9.8 </em><em />
<em />
Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity, <em>vsin ( θ )</em> , and not the horizontal component, <em>vcos ( θ )</em> .
Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: <em>vsin
(
θ
) =
</em>
Where <em> vsin ( θ )</em> is the vertical component of the initial velocity and <em>h</em> and <em>g</em> are max height and gravitational acceleration constant respectively.
Plugin, rearrange and solve
v
sin
(
θ
) =
v sin ( 45 ∘ ) = √ 2 × 3.7 × 9.8
v
(
0.71
) =
v ( 0.71 ) = 8.52
v = 8.52 /0.71
v = 12 m s
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Complete Question
A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.
Answer:
The phase at the second point is
Explanation:
From the question we are told that
The wavelength of the spherical wave is
The first radius is
The phase at that instant is
The second radius is
Generally the phase difference is mathematically represented as
this can also be expressed as
So we have that
substituting values