<h2>
Angular acceleration is 80 rad/s²
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Number of revolutions undergone is 1.02</h2>
Explanation:
We have equation of motion v = u + at
Initial angular velocity, u = 0 rad/s
Final angular velocity, v = 32 rad/s
Time, t = 0.40 s
Substituting
v = u + at
32 = 0 + a x 0.40
a = 80 rad/s²
Angular acceleration is 80 rad/s²
We have equation of motion s = ut + 0.5 at²
Initial angular velocity, u = 0 rad/s
Angular acceleration, a = 80 rad/s²
Time, t = 0.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.4 + 0.5 x 80 x 0.4²
s = 6.4 rad
Angular displacement = 6.4 rad
Number of revolutions undergone is 1.02
“The mechanical energy is conserved" in the given system is true out of all given options.
Answer: Option 2
<u>Explanation:
</u>
According to law of conservation of energy, the energy will neither be created nor be destroyed, irrespective of the type of energy. As in the present case, the ball is bowled and it is travelling to ground, so the mechanical energy is working in this case. Thus the mechanical energy will be conserved. Even it can be shown as follows.
As the 2 kg ball is travelling with a speed of 7 m/s, the kinetic energy exhibited by the ball while falling to ground will be
Thus, applying given values, we get,
Similarly, as the ball is 2.5 m above ground, the potential energy will also be exhibited by the ball at that position. So the potential energy will be
Thus,
Thus as the magnitude of kinetic energy is equal to the magnitude of potential energy exhibited by the ball with varying direction, the net energy will be zero. This is because the kinetic energy will be acting in opposite direction to the potential energy exhibited by the ball. Hence as the net energy is zero, the mechanical energy is conserved.
For this case, the first thing you should do is define a reference system.
Once the system is defined, we must follow the following steps:
1) Do the sum of forces in a horizontal direction
2) Do the sum of forces in vertical direction
The forces will be balanced if for each direction the net force is equal to zero.
The forces will be unbalanced if for each direction the net force is nonzero.
Answer:
Add the forces in the horizontal and vertical directions separately.
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)