A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons
1 answer:
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Answer:
The radius is decreasing at 4 mm/s
Explanation:
The volume of a sphere is:
So, when the volume is 972π mm^3 the radius r is:
r = 9mm
Now, the change rate is given by the derivative:
Where: dV/dt = -324π mm^2/s
r = 9mm
Solving for dr/dt:
dr/dt = -4mm/s
Explanation:
It is given that,
Diameter of the semicircle, d = 45 m
Radius of the semicircle, r = 22.5 m
Speed of greyhound, v = 15 m/s
The greyhound is moving under the action of centripetal acceleration. Its formula is given by :
We know that,
Hence, this is the required solution.
Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then
We know that the relation between the tangential velocity with the angular velocity is given by,
Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity
Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
Therefore teh average angular acceleration of each wheel is