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3 years ago

Which diagram best shows the results of removing heat from the original sample until it freezes?

Physics

1 answer:

vova2212 [387]3 years ago

5 0

Answer:

Explanation:

the closest molecules are in b witch would be a solid

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Dynamics Newton's 1st & 2nd Laws

pav-90 [236]

Ummm what does this even mean I need a picture to help u

8 0

3 years ago

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

8 0

4 years ago

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

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

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

Scientific theories have strod the test of time and will not change

Nataly [62]

That is not a question but not all scientific theories have stood the test of time

8 0

4 years ago

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

$\displaystyle{\int_{x_i}^{x_j}F(x)dx}$

and substituting with the function of our problem:

$\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}$

5 0

4 years ago