Answer:
19.9 N/m
Explanation:
From the question,
Applying Hook's law
F = Ke.................. Equation 1
Where F = Force on the spring, k = spring constant, e = extension
But the force on the spring is the weight of the mass
Therefore,
mg = ke.................. Equation 2
Where m = mass. g = acceleration due to gravity
make e the subject of the equation
e = mg/e................ Equation 3
Given: m = 455 g = 0.455 kg, e = 22.4 cm = 0.224 m,
Constant: g = 9.8 m/s²
Substitute these values into equation 3
e = (0.455×9.8)/0.224
e = 19.9 N/m
Density can be calculated using the following rule:
density = mass / volume
Since the density is given as 8 gm/cm^3 and the mass is given as 600 grams, therefore, all we need to do is substitute in the equation to get the value of the volume as follows:
8 = 600 / volume
volume = 600 / 8 = 75 cm^3
Making prediction , hypothesis and guessing to see if it’s right , and leading
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
Answer:
t / τ₀ = 2.9
the time constant has to be multiplied by approximately three
Explanation:
The expression for long of a capacitor is
q = q₀ (1- e (-t/τ₀)
Where q is the charge and τ₀ is the time constant. What is found for when the initial capacitor charge at 63.5% (e⁻¹) of the maximum charge
In our case we are asked that the load be 94.5%, this means that
q/q₀ = 94.5%
q/q₀ = 0.945
Let's replace and calculate
q/q₀ = 1 - e (-t/τ₀)
0.945 = 1 - e (-t/τ₀)
(-t /τ₀) = ln (1 -0.945) = ln (0.055)
t / τ₀ = 2.9
the time constant has to be multiplied by approximately three for this charge
