Can you pickle a pickle? Like take a cucumber that's already pickled, can you then pickle that pickled cucumber?

joja [24]

If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is not heated.

8 0

3 years ago

Read 2 more answers

Consider two diffraction gratings. One grating has 3000 lines per cm, and the other one has 6000 lines per cm. Both gratings are

Usimov [2.4K]

Answer:

The 6000 lines per cm grating, will produces the greater dispersion .

Explanation:

A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.

The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.

For a plane diffraction grating, the angular positions of principle maxima is given by

(a + b) sin ∅n = nλ

where

a+b is the distance between two consecutive slits

n is the order of principal maxima

λ is the wavelength of the light

From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.

In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.

7 0

3 years ago

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

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

Here,

L = Length

g = Acceleration due to gravity

We can realize that $2 \pi$ is a constant so it is proportional to the square root of its length over its gravity,

$T \propto \sqrt{\frac{L}{g}}$

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

$g \rightarrow 0 \Rightarrow T \rightarrow \infty$

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0

3 years ago

Hammer falls off a roof top and strikes the ground with a certain kinetic energy. If it fell

Galina-37 [17]

Answer:

If the starting GPE is doubled than it's KE would also double.

6 0

3 years ago

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

So

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

So

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

3 years ago