Answer:
a. Springs oscillate with the same frequency
Explanation:
As they both are in the same height at equilibrium, so
weight of ball must be balanced with spring force, that is
k×x=mg
k= stiffness constant of spring
x=distance stretched
g= acceleration due to gravity
so, we can write
k/m=g/x
as the g is a constant and they stretched to same distance x so the g/x term becomes constant and
and k/m is same for both the springs so they will oscillate at the same frequency.
hence option a is correct.
Answer:
d. less than 20m/s
Explanation:
To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is
[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s
As 14.14m/s is less than 20m/s. d is the correct selection for this question.
Answer:
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
Explanation:
Acceleration is the change in velocity per unit time
a = ∆v/t
Given;
∆v = 50.0miles/hour - 0
∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour
∆v = 22.352m/s
t = 2.22 s
So,
Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
Answer: 1. decreasing the mass of both objects
2. decreasing the mass of one of the objects
3. increasing the distance between the objects
Explanation: Hope that helped! (:
Answer:
ω = √(2T / (mL))
Explanation:
(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.
The x-components of the tension forces cancel each other out, so the net force is in the y direction:
∑F = -2T sin θ, where θ is the angle from the horizontal.
For small angles, sin θ ≈ tan θ.
∑F = -2T tan θ
∑F = -2T (Δy / L)
(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)
In this case, k = 2T/L, so the frequency is:
ω = √((2T/L) / m)
ω = √(2T / (mL))
