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3 years ago

Imma give u all my points and ill make u brainlist jus answer them

Physics

1 answer:

Reptile [31]3 years ago

3 0

I'll put my answer in the comments but, part of the question is cut off do you mind putting a more readable and not sideways picture?

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dam is used to block the passage of a river and to generate electricity. Approximately 58.4 x 103 kg of water falls each second

mrs_skeptik [129]

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep = 58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

3 0

3 years ago

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

7 0

3 years ago

Which of the following energies defines energy stored inside a spring?

Leya [2.2K]

Answer:

Elastic potential energy

7 0

2 years ago

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0

2 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers

Imma give u all my points and ill make u brainlist jus answer them​

Imma give u all my points and ill make u brainlist jus answer them