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Darya [45]
3 years ago
12

The rate of change of the moment of angular momentum of the fluid flowing through the produced impeller of a turbine is equal to

the a. Work b. Power c. Torque d. Thrust
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

Option (c)

Explanation:

The rate of change of linear momentum is called force.

As the linear motion terms are analogous to the terms in rotational motion.

So, the rate of change of angular momentum is called torque.

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A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precise
Alex73 [517]
<h2>Ratio of free fall acceleration of Tokyo to Cambridge = 0.998</h2>

Explanation:

We know the equation

            T=2\pi \sqrt{\frac{l}{g}}

   where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

              g= \frac{4\pi^2l}{T^2}

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

                     \frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

6 0
3 years ago
The specific heat of a substance is the energy required to produce a certain change in _____________. A. appearance B. volume C.
CaHeK987 [17]
I think C I’m not 100% sure.
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No matter what values of m and k you used for the spring, what is the ratio of the period to k m.
Leto [7]

The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to \sqrt{\dfrac{m}{k} }.

Response:

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

<u />

<h3>How is the value of the ratio of the period to \sqrt{\dfrac{m}{k} } calculated?</h3>

Given:

The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the

mass attached to the spring <em>m</em> is presented as follows;

T =  \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}

Therefore, the fraction of of the period to \sqrt{\dfrac{m}{k} }, is given as follows;

\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi

2·π ≈ 6.23

Therefore;

T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1

Which gives;

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

Learn more about the oscillations in spring here:

brainly.com/question/14510622

7 0
3 years ago
High rate of speed?
defon

<em>velocity over time I think</em>

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