Answer:
a. 2 Hz b. 0.5 cycles c . 0 V
Explanation:
a. What is period of armature?
Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz
b. How many cycles are completed in T/2 sec?
The period, T = 1/f = 1/2 Hz = 0.5 s.
So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,
Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.
c. What is the maximum emf produced when the armature completes 180° rotation?
Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0
E = E₀ × 0 = 0
E = 0
So, at 180° rotation, the maximum emf produced is 0 V.
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
sir what's the question you have all you wrote is it's not b
Answer:
a

b

Explanation:
From the question we are told that
The pressure of the water in the pipe is
The speed of the water is 
The original area of the pipe is
The new area of the pipe is
Generally the continuity equation is mathematically represented as

Here
is the new velocity
So

=> 
=> 
=> 
=> 
Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure
This is mathematically represented as

Here
is the density of water with value
![P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20P_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Crho%20%5B%20v_1%5E2%20-%20v_2%5E2%20%5D)
=> ![P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20110%20%2A10%5E%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%201000%20%2A%20%20%5B%201.4%20%5E2%20-%205.6%20%5E2%20%5D)
=> 
Answer:
The strong person should carry the ladder at the front end and the weak person should carry it at the back end.
Explanation:
this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.
However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object
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<h2><em>I Hope it help you </em></h2>