A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
1 answer:
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Answer:
a) The sign of the charge is positive.
b) The magnetic force on the particle is 0.050 newtons.
Explanation:
The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:
(1)
a)
Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.
b)
Now by 1:
The initial volume of the gas is
while its final volume is
so its variation of volume is
The pressure is constant, and it is
Therefore the work done by the gas is
where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is
And the change in internal energy of the gas can be found by using the first law of thermodynamics:
where the positive sign means the internal energy of the gas has increased.
while its final volume is
so its variation of volume is
The pressure is constant, and it is
Therefore the work done by the gas is
where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is
And the change in internal energy of the gas can be found by using the first law of thermodynamics:
where the positive sign means the internal energy of the gas has increased.
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
