Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = 
x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Answer:
The change in angular momentum is 
Explanation:
From the question we are told that
The initial angular velocity of the spinning top is 
The angular velocity after it slow down is 
The time for it to slow down is 
The rotational inertia due to friction is 
Generally the change in the angular momentum is mathematically represented as
substituting values
Answer:
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Explanation:
Total force required = Mass x Acceleration,
F = ma
Here we need to consider the system as combine, total mass need to be considered.
Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg
We need to accelerate the group of rocks from the road at 0.250 m/s²
That is acceleration, a = 0.250 m/s²
Force required, F = ma = 1744 x 0.25 = 436 N
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Answer:
The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)
Explanation:
we have to:
m = mass = 4 kg
k = spring constant = 577 N/m
c = damping constant = 4 N*s/m
The differential equation of motion is equal to:
m(d^2x/dt^2) + c(dx/dt) + k*x = 0
Replacing values:
4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0
Thus, we have:
4*x^2 + 4*x + 577 = 0
we will use the quadratic equation to solve the expression:
x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2) ± 12i
The solution is equal to:
x = e^(1/2) * (c1*sin12t + c2*cos12t)
x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)
We have the follow:
x(0) = 5
e^0(0*c1 + c2) = 5
c2 = 5
x´(0) = 7
(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7
(-1/2)*(5) + 12*c1 = 7
Clearing c1:
c1 = 0.792
The function is equal to:
x = e^(-t/2) * (0.792*sin12t + 5cos12t)
Answer:
The current through the wire is equal to 0.8 A.
Explanation:
Given that,
The length of a copper wire = 2 m
Potential difference = 24 mV
The current through the wire is 0.40 A.
The new potential difference is 48 mV.
We need to find the current through the wire.
As the potential difference is doubled for second wire. So the new current will be :
I' = 2I
= 2 × 0.4
= 0.80 A
So, the current through the wire is equal to 0.8 A.
