Answer:
a) F = 2.7 10⁻¹⁴ N
, b) a = 2.97 10¹⁶ m / s² c) θ = 14º
Explanation:
The magnetic force on the electron is given by the expression
F = q v x B
Which can be written in the form of magnitude and the angle found by the rule of the right hand
F = q v B sin θ
where θ is the angle between the velocity and the magnetic field
a) the maximum magnitude of the force occurs when the velocity and the field are perpendicular, therefore, without 90 = 1
F = e v B
F = 1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²
F = 2.73 10⁻¹⁴ N
F = 2.7 10⁻¹⁴ N
b) Let's use Newton's second law
F = m a
a = F / m
a = 2.7 10⁻¹⁴ / 9.1 10⁻³¹
a = 2.97 10¹⁶ m / s²
The actual acceleration (a1) is a quarter of this maximum
a1 = ¼ a
a1 = 7.4 10¹⁵ m / s²
With this acceleration I calculate the force that is executed on the electron
F = ma
e v b sin θ= ma
sin θ = ma / (e v B)
sin θ = 9.1 10⁻³¹ 7.4 10¹⁵ / (1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²)
sin θ = 6.734 10⁻¹⁵ / 27.26 10⁻¹⁵
sin θ = 0.2470
θ = 14.3º
Answer:
30m/s
Explanation:
From law of motion equation
Vf= Vi + at
Where Vf= final velocity
Vi= initial velocity=0(the car started at rest)
a= acceleration= 3m/s2
t= time= 10s
Then substitute into the equation to get the final velocity.
Vf= 0+(10×3)
Vf= 30m/s
Hence, the car's final velocity is 30m/s
Answer:
Explanation:
Given that
Intensity I
Radius of earth,R = 6370 Km
We know that surface area of earth, A
As we know that pressure due to intensity given as
V =Velocity of light
We know that force F
F = P .A
b)Gravitational force F
So F
<h2>
Answer: zero (0)</h2>
Explanation:
The orbit of a body around another in space, is described by six orbital elements that determine its orientation, position, size and shape.
In the specific case of the shape of the orbit, this is determined by its <u>eccentricity</u>, which varies between 0 and 1 in the case of closed orbits (circle and ellipse). When the eccentricity is 0, the shape of the orbit is circular, when this value begins to vary until approaching 1 (without reaching 1), the shape of the orbit becomes more elliptical.
In this sense, a circular orbit will have an eccentriciy of zero.