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3 years ago

10

Under what condition is it possible for liquid water to have a higher temperature than water vapor when both are in the same con

tainer?

1 answer:

ratelena [41]3 years ago

3 0

Answer:

See there is no condition your question is so true.

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A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

$\omega^2 = 5.77$

$\omega = 2.4 rad/s$

so the ratationa frequency is given by

$\omega = 2 \pi f$

$2.4 = 2 \pi f$

$f = \frac{2.4}{2\pi}$

$f = 0.38 Hz$

7 0

3 years ago

What must happen to the temperature of a material for thermal expansion to occur.

Aloiza [94]

In most cases the temperature must increase for thermal expansion to occur. Most substances expand as temperature increases because the atoms or molecules vibrate faster as temperature increases and experience greater separation.

8 0

3 years ago

if a kid is practicing for a play and there is an auditorium of loud people that are shouting, and he screams, he hears and echo

Vilka [71]

Because the sound waves are being observed by the people that are there and it is not empty as in an empty place the sound waves will come back to you.

5 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

Suppose a pendulum bob is swinging back and forth. If the highest point is 12.5 cm above the lowest point, what would be speed o

Colt1911 [192]

Answer:

The speed of the bob when it passes the lowest point $V = 1.57 \frac{m}{s}$

Explanation:

Given data

$H = 12.5 \ cm = 0.125 \ m\\$

When a pendulum swinging back & forth then at highest point the velocity is zero and lowest point velocity is maximum.

Velocity at lowest point is given by

$V = \sqrt{2 g H}$

$V = \sqrt{2 (9.81)(0.125)}$

$V = 1.57 \frac{m}{s}$

Therefore the speed of the bob when it passes the lowest point $V = 1.57 \frac{m}{s}$

7 0

3 years ago