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3 years ago

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Much of the energy of falling water in a waterfall is converted into heat. If all the mechanical energy is converted into heat t

hat stays in the water, how much of a rise in temperature occurs in a 111 m waterfall?

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

Explanation:

potential energy of water is converted into heat energy

P E = mgh = m x 9.8 x 111

= 1087.8 m .

Let rise in temperature θ

m x s x θ = heat absorbed

m x s x θ = 1087.8 m

m x 4182 x θ = 1087.8 m

θ = 0.26 degree celsius .

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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

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brainly.com/question/2438000

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The correct answer to this question is 15

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2 years ago

during convection, hot material _____ (expands & sinks, cools & rises, expands & rises, deflates & rises) then m

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Answer:

During convection, hot material expands & rises then moves to the side and cools & sinks. this circular pattern is called a convection current.

Explanation:

Convection is one of the three methods of transfer of heat. It occurs only in fluids (liquids or gases).

Convection occurs when there is a source of heat that heats a fluid, such as in a boiling pot of water. The water which is on the bottom of the pot becomes warmer before than the water at the top (because it is closer to the flame), and so it becomes less dense: for this reason, it expands and it becomes rising. On the contrary, the water on top is colder, so it is more dense and starts sinking, replacing the warmer water. As the new part of water gets warmer, it starts rising, and so the process is continuously repeated. This circular current is called convection current.

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3 years ago

Read 2 more answers

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

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Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
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Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

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Now u is the final velocity of the ball with respect to the top of the building

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$3.96=gt\\t=0.4\ \rm s\\$

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3 years ago