Answer:
Explanation:
I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.
First row, KE = 0
Second row, KE = 220500 J
Third row, KE = 183750 J
Fourth row, KE = 205800 J
That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.
Answer:
Explanation:
The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them. ... The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them.
Because the direction of the kicks are opposite, the net force between the applied forces is their difference.
Fn = F₂ - F₁
Substituting,
Fn = 15 N - 5 N
Fn = 10 N
From Newton's second law of motion,
Fn = m x a
where m is mass and a is acceleration. Manipulating the equation so that we are able to calculate for a,
a = Fn / m
Substituting,
a = (10 N) / 2 kg
a = 5 m/s²
<em>ANSWER: 5 m/s²</em>
Answer:
The ball takes 5s to reach the ground
Explanation:
in order to solve this problem we use the kinematics equation with gravity as acceleration:
![h=v_0t+1/2*gt^2](https://tex.z-dn.net/?f=h%3Dv_0t%2B1%2F2%2Agt%5E2)
we replace the values
![0=1/2*32t^2+16t-480](https://tex.z-dn.net/?f=0%3D1%2F2%2A32t%5E2%2B16t-480)
We solve this quadratic equation:
t=5s
t=-6s (this solution has not physical sense)
Answer:
The mass of the astronaut is approximately 119.74 kg
Explanation:
Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:
(1)
First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:
<em>(it’s important to note that one complete vibration is the period of the movement)</em>
![1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2}](https://tex.z-dn.net/?f=1.3%3D2%5Cpi%5Csqrt%7B%5Cfrac%7B42.5%7D%7Bk%7D%7D%5CLongrightarrow%20k%3D42.5%28%5Cfrac%7B2%5Cpi%7D%7B1.3%7D%29%5E%7B2%7D%20)
![k\approx992.8\,\frac{N}{m}](https://tex.z-dn.net/?f=k%5Capprox992.8%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20)
Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds
![2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5](https://tex.z-dn.net/?f=%202.54%3D2%5Cpi%5Csqrt%7B%5Cfrac%7B42.5%2BM_%7Ba%7D%7D%7B992.8%7D%7D%5CLongrightarrow%20M_%7Ba%7D%3D992.8%28%5Cfrac%7B2.54%7D%7B2%5Cpi%7D%29%5E%7B2%7D-42.5%20)
![\mathbf{M_{a}\approx119.74\,kg}](https://tex.z-dn.net/?f=%20%5Cmathbf%7BM_%7Ba%7D%5Capprox119.74%5C%2Ckg%7D%20)