.A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30 angle to the horizontal. The forc
1 answer:
To solve this problem, it is necessary to apply the concepts related to Work according to the Force and distance, as well as the concepts related to energy lost-or gained-by heat. Mathematically the energy corresponding to heat is given as:
Where,
m = mass
= Specific heat
= Change in Temperature
At the same time the Work made by the Force and the distance is given as:
As the force is applied at an angle of 30 degrees, the efficient component would be given by the vertical then the work / energy would be determined as:
Now this energy is used to heat the aluminum. We can find the change at the temperature as follow:
Therefore the correct answer is B.
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Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
A)
B)
C)
D)
E)
Explanation:
A)
When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:
where
q is the charge of the particle (positive)
On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):
where
m is the mass of the particle
v is its final speed
According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:
And solving for v, we find the speed v at which the particle enters the cyclotron:
B)
When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is
where B is the strength of the magnetic field.
This force acts as centripetal force, so we can write:
where r is the radius of the orbit.
Since the two forces are equal, we can equate them:
And solving for r, we find the radius of the orbit:
(1)
C)
The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.
It can be calculated as the ratio between the length of the circumference () and the velocity of the particle (v):
(2)
From eq.(1), we can rewrite the velocity of the particle as
Substituting into(2), we can rewrite the period of revolution of the particle as:
And we see that this period is indepedent on the velocity.
D)
The angular frequency of a particle in circular motion is related to the period by the formula
(3)
where T is the period.
The period has been found in part C:
Therefore, substituting into (3), we find an expression for the angular frequency of motion:
And we see that also the angular frequency does not depend on the velocity.
E)
For this part, we use again the relationship found in part B:
which can be rewritten as
(4)
The kinetic energy of the particle is written as
So, from this we can find another expression for the velocity:
And substitutin into (4), we find:
So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.
Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).
Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
The drag force on the athlete is 33 N