Answer:
Workdone = 1.96 Nm
Explanation:
Given the following data;
Mass = 100g to kg 100/1000 = 0.1Kg
Distance = 2m
To find the workdone;
First of all, we would determine the force acting on the apple
Force = mass * acceleration due to gravity
But, acceleration due to gravity is equal to 9.8m/s²
Force = 0.1 * 9.8
Force = 0.98N
Now, we can find the workdone using the formula;
Workdone = force* distance
Substituting into the equation, we have
Workdone = 0.98 * 2
Workdone = 1.96 Nm
Answer:
The frequencies are 13.8 Hz, 75 Hz, 12 Hz and 63.8 Hz.
Explanation:
Given that,
The frequency in r.p.m
![f_{1}=830\ rpm](https://tex.z-dn.net/?f=f_%7B1%7D%3D830%5C%20rpm)
![f_{2}=4500\ r.p.m](https://tex.z-dn.net/?f=f_%7B2%7D%3D4500%5C%20r.p.m)
![f_{3}=720\ r.p.m](https://tex.z-dn.net/?f=f_%7B3%7D%3D720%5C%20r.p.m)
![f_{4}=3832\ r.p.m](https://tex.z-dn.net/?f=f_%7B4%7D%3D3832%5C%20r.p.m)
Suppose, we find the frequency in hz.
We know that,
One r.p.m is equal to the one divided by 60 Hz.
![1\ r.p.m=\dfrac{1}{60}\ Hz](https://tex.z-dn.net/?f=1%5C%20r.p.m%3D%5Cdfrac%7B1%7D%7B60%7D%5C%20Hz)
We need to calculate the frequency in Hz
Using formula for frequency in Hz
For f₁,
![f_{1}=830\ r.p.m](https://tex.z-dn.net/?f=f_%7B1%7D%3D830%5C%20r.p.m)
![f_{1}=\dfrac{830}{60}\ Hz](https://tex.z-dn.net/?f=f_%7B1%7D%3D%5Cdfrac%7B830%7D%7B60%7D%5C%20Hz)
![f_{1}=13.8\ Hz](https://tex.z-dn.net/?f=f_%7B1%7D%3D13.8%5C%20Hz)
For f₂,
![f_{2}=4500\ r.p.m](https://tex.z-dn.net/?f=f_%7B2%7D%3D4500%5C%20r.p.m)
![f_{2}=\dfrac{4500}{60}\ hz](https://tex.z-dn.net/?f=f_%7B2%7D%3D%5Cdfrac%7B4500%7D%7B60%7D%5C%20hz)
![f_{2}=75\ Hz](https://tex.z-dn.net/?f=f_%7B2%7D%3D75%5C%20Hz)
For f₃,
![f_{3}=720\ r.p.m](https://tex.z-dn.net/?f=f_%7B3%7D%3D720%5C%20r.p.m)
![f_{3}=\dfrac{720}{60}\ hz](https://tex.z-dn.net/?f=f_%7B3%7D%3D%5Cdfrac%7B720%7D%7B60%7D%5C%20hz)
![f_{3}=12\ Hz](https://tex.z-dn.net/?f=f_%7B3%7D%3D12%5C%20Hz)
For f₄,
![f_{4}=3832\ r.p.m](https://tex.z-dn.net/?f=f_%7B4%7D%3D3832%5C%20r.p.m)
![f_{4}=\dfrac{3832}{60}\ hz](https://tex.z-dn.net/?f=f_%7B4%7D%3D%5Cdfrac%7B3832%7D%7B60%7D%5C%20hz)
![f_{4}=63.8\ Hz](https://tex.z-dn.net/?f=f_%7B4%7D%3D63.8%5C%20Hz)
Hence, The frequencies are 13.8 Hz, 75 Hz, 12 Hz and 63.8 Hz.
Answer:
Approximately 3.03 seconds.
Explanation:
The distance traveled in the vertical direction is given by the kinematic equation:
![\displaystyle y = v_{iy}t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20v_%7Biy%7Dt%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
Where <em>v</em>_<em>iy</em> and <em>a</em> are the initial velocity and acceleration of the object, respectively, in the vertical direction.
Because the rock is thrown horizontally, there is no horizontal velocity. Therefore:
![\displaystyle y = \frac{1}{2} at^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
The vertical acceleration is simply gravity <em>g. </em>This, this yields the general equation:
![\displaystyle y = \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substitute 45 m for <em>y</em> and solve for time <em>t:
</em>![\displaystyle \begin{aligned} (45\text{ m}) & = \frac{1}{2}(9.8\text{ m/s$^2$})t^2 \\ \\ t^2 & =\frac{450}{49}\text{ s$^2$} \\ \\ & \approx 3.03\text{ s}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%2845%5Ctext%7B%20m%7D%29%20%26%20%3D%20%5Cfrac%7B1%7D%7B2%7D%289.8%5Ctext%7B%20m%2Fs%24%5E2%24%7D%29t%5E2%20%5C%5C%20%5C%5C%20t%5E2%20%26%20%3D%5Cfrac%7B450%7D%7B49%7D%5Ctext%7B%20s%24%5E2%24%7D%20%20%5C%5C%20%5C%5C%20%20%26%20%5Capprox%203.03%5Ctext%7B%20s%7D%5Cend%7Baligned%7D)
Therefore, it will take approximately 3.03 seconds for the rock to fall 45 meters vertically.
The work done by the engine is equal to the gravitational potential energy (GPE) put into it. GPE = mgh
GPE = Work = 1200·9.81·13 = 153,036J
The definition of dilute is "make (a liquid) thinner or weaker by adding water or another solvent to it." Now, this may make you think that the beaker with three scoops is the most dilute, but it's not. In this case, it depends on the salt to water ratio. Let's say each beaker contains five parts water. The first beaker has a ratio of 1/5. The second had a ratio of 2/5. The third has a ratio of 3/5. To find which has the most water compared to the others, I'll use equal to make the numerator (The amount of salt) seemingly equal each time. Just a warning, this strategy doesn't work every time. Now, if we make the numerators the same, that means which ever denominator is the highest will be the most dilute solution. Let's make each numerator equal to six, as each number (1, 2, and 3) go into six.
1/5 = 6/30
2/5 = 6/15
3/5 = 6/12
I got these numbers by dividing six (What we want the numerator to be) by each current numerator, and then multiplying the quotient (The answer of a division problem) by both sides of the fraction. Since the first beaker has the highest denominator, we know that it is the most dilute.
mark brainliest ;)