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2 years ago

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Physics

1 answer:

Effectus [21]2 years ago

8 0

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

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Which quantity is most closely related to the speed of gas particles

natali 33 [55]

Temperature is a measure of the energy of molecules and energy is related to speed.

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<span>Oxygen will form ionic bonds with nitrogen. The others will not by themselves. </span>

<span>A high pH is indicative of a basic solution. HCl and H2SO4 are both strong acids and will result in a lower (more acidic) pH. Water is the standard. KOH is a strong base and will increase the pH. </span>

<span>An acidic solid will lower the pH of a solution. pH measures the number of hydrogen ions and a lower pH will mean that there are more H+ ions, but the solid could just have reacted with the OH- to cause the pH to decrease</span>

3 0

2 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago

If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the

Nataliya [291]

The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

= 10 gram

Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.

Learn more about weight on:

brainly.com/question/19753744

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1 year ago

When the Earth is closest to the Sun, it is at:

ahrayia [7]

a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
We pass it every year early in January.

b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
We pass it every year early in July.

c). Proxihelion . . . a made-up, meaningless word

d). Equinox . . . the points on the map of the stars where the Sun
appears to be on March 21 and September 21.

5 0

3 years ago

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$