Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:im just guessing d but i think its d though
Explanation:
it pretty obvious
The right answer for the question that is being asked and shown above is that: "C) The alpha particle will be deflected in a curve path. " the result of an alpha particle coming into a magnetic field is that <span>C) The alpha particle will be deflected in a curve path. </span>
Work = (force) x (distance)
The worker does (40N) x (4m) = 160 joules of work.
Friction eats up (27N) x (4m) = 108 joules of that energy,
generating 108 joules of heat.
The remaining (160J - 108J) = 52 joules of energy moves the box.
