Answer:
a
The number of radians turned by the wheel in 2s is 
b
The angular acceleration is 
Explanation:
The angular velocity is given as
Now generally the integral of angular velocity gives angular displacement
So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec
This is mathematically evaluated as
![= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.](https://tex.z-dn.net/?f=%3D%20%5B%5Cfrac%7B2t%5E2%7D%7B2%7D%20%2B%20%5Cfrac%7Bt%5E4%7D%7B4%7D%5D%20%5Cleft%5C%7B%202%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0](https://tex.z-dn.net/?f=%3D%20%5B%5Cfrac%7B2%282%5E2%29%7D%7B2%7D%20%2B%20%5Cfrac%7B2%5E4%7D%7B4%7D%5D%20-%200)
Now generally the derivative of angular velocity gives angular acceleration
So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration
This is mathematically evaluated as
so at t=2
Answer:
1.5min
Explanation:
To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.
By definition centripetal acceleration is given by
Where,
V = Tangencial velocity
r = radius
With our values we know that
Therefore solving to find V, we have:
For definition we know that the Time to complete are revolution is given by
Answer: D. They are the coldest stars.
Explanation:
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;
<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>
Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
Learn more here:brainly.com/question/15345295
Answer:
net force would be 50 N right
Explanation:
