Ask question

Ask question

All categories

3 years ago

13

What is the usual time ice takes to evaporate above a Bunsen burner?

1 answer:

Step2247 [10]3 years ago

5 0

Depends on the volume but a normal ice cube takes less than 20 seconds

You might be interested in

An air conditioner operating between 93°F and 70°F is rated at 4000 Btu/h cooling capacity. Its coefficient of performance is 27

lana [24]

Answer: $P_{in}=0.25\ Hp$

Explanation:

Given

Operating temperatures are

$T_L=70^{\circ}F\approx 294.261\ K$

$T_H=93^{\circ}F\approx 307\ K$

$COP$ is $27\%$ of Carnot refrigerator

$COP_{carnot}=\dfrac{T_L}{T_H-T_L}$

$COP_{carnot}=\dfrac{294.261}{307-294.261}$

$COP_{carnot}=\dfrac{294.261}{12.74}$

$COP_{carnot}=23.09\approx 23.1$

actual $COP=0.27 \times COP_{carnot}$

$COP_{actual}=6.237$

Also, $COP=\dfrac{\text{Desired effect}}{\text{Power in}}$

$6.237=\frac{4000}{P_{in}}$

$P_{in}=\frac{4000}{6.237}=641.33\ Btu/h$

$P_{in}=0.25\ Hp$

6 0

3 years ago

The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating havin

nadya68 [22]

Answer:

The range of angles is from 17.50° to 31.76°

Explanation:

The diffraction grid equation is as follows:

$dsen\theta=m\lambda$

Clearing for $\theta$

$sen\theta=\frac{m\lambda}{d}$

$\theta=sen^{-1}(\frac{m\lambda}{d})$

where $\theta$ is the angle, $m$ is the order, in this case $m=1$ , $\lambda$ is the wavelength, and $d$ is defined as follows:

$d=\frac{1}{resolution}$

and since the resolution is 750 lines/mm wich is the same as $750lines/1x10^{-3}m$

$d$ will be:

$d=\frac{1}{750lines/1x10^{-3}m}=\frac{1x10^{-3}m}{750lines}=1.33x10^{-6}m$

wich is the distance between each line of the diffraction grating.

substituting the values for $m$ and $d$ :

$\theta=sen^{-1}(\frac{(1)\lambda}{(1.33x10^{-6}m)})$

And we need to find two angle values: one for when the wavelength is 400nm and one for when it is 700 nm. So we will get the angle range

$\theta=sen^{-1}(\frac{(400x10^{-9})}{(1.33x10^{-6}m)})=17.50$

and

$\theta=sen^{-1}(\frac{(700x10^{-9})}{(1.33x10^{-6}m)})=31.76$

The range of angles is from 17.50° to 31.76°

3 0

3 years ago

If an object is 413 Kg in mass and 640 M3. What is the density? Will It Float?

monitta

Answer:

The density will be 0.65 kg/m3

I think it won't float!! I am not sure though.

7 0

3 years ago

a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th

Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

And solving for $F_f$ , we find the force of friction:

$F_f = F-ma=500-(50)(4)=300 N$

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

4 0

3 years ago

A book weighing 12 n is placed on a table. how much support force does a table exert on the book?

Alenkasestr [34]

It should be 12 N. the force of the book on the table should be the same as the force of the table on the book.

7 0

4 years ago