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Andru [333]
3 years ago
8

A turbine of a coal burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0

% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1560 V, what current does it deliver (in A)?
Physics
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

I=573.846\ A

Explanation:

Given:

  • mechanical energy delivered to the generator, E_m=1500\ hp=1119000\ W
  • efficiency of the generator in converting the mechanical energy into electrical, \eta_g=80\%
  • terminal potential difference across the generator output, V=1560\ V

<u>Now the amount of energy converted into electrical energy by the generator:</u>

E_e=E_m\times \eta_g

E_e=1119000\times 80\%

E_e=895200\ W

As power of current is given by:

P=V.I

where:

I= current

Now,

E_e=P

895200=V.I

I=\frac{895200}{1560}

I=573.846\ A

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Vector quantity is defined by direction as well as magnitude both

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2) Speed = it is ratio of total distance covered and total time, so it also do not require any direction.

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4) time = time is the measurement of the interval of two events and we do not require any direction in it so its a scalar quantity.

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3 years ago
A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
seropon [69]

Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

or

\theta=50\ revolution

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0
3 years ago
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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