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3 years ago

What problems would we have if Pascal had failed to give the Pascal's law? Write some points.

2 answers:

4 0

Hydraulic jacks, automobile brakes and even the lift generated on airplane wings can be explained using Pascal's principle. Pascal's principle is based on the idea that fluids at rest are incompressible, allowing very large forces to be transmitted with the application of a smaller force.

Gemiola [76]3 years ago

4 0

Answer:

Pascal's law states that a change in pressure at any point in an enclosed fluid is transmitted equally throughout the fluid.

Explanation:

Applications of Pascal's law include

>The hidden standard of the water powered jack and pressure driven press.

>Power intensification in the slowing mechanism of most engine vehicles.

>Utilized in artesian wells, water pinnacles, and dams.

>Scuba jumpers must comprehend this rule. At a profundity of 10 meters submerged, pressure is double the environmental weight adrift level, and increments by around 100 kPa for each expansion of 10 m depth.[5]

>Generally Pascal's standard is applied to bound space (static flow), however because of the consistent flow process, Pascal's rule can be applied to the lift oil instrument (which can be represented as a U tube with pistons on either end).

Without Pascal's law, none of the above could be possible.

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A simple pendulum consists of a 2 kg bob attached to a 1.5 m long string. How much time (in s) is required for this pendulum to

Charra [1.4K]

Answer:

6.15 s

Explanation:

The period of a simple pendulum is given by the equation

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

where

L is the length of the pendulum

g is the acceleration of gravity

For the pendulum in this problem,

L = 1.5 m (length)

$g=9.8 m/s^2$ (acceleration due to gravity on Earth)

Therefore, its period is

$T=2\pi \sqrt{\frac{1.5}{9.8}}=2.46 s$

And therefore, the time taken for the pendulum to complete 2.5 oscillations is equal to 2.5 times the period:

$t=2.5T=(2.5)(2.46)=6.15 s$

3 0

3 years ago

A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmoni

SVETLANKA909090 [29]

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

$E=\dfrac{1}{2}kA^2$

Put all the values,

$E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J$

So, the value of total mechanical energy is equal to 0.01 J.

3 0

3 years ago

- Atoms of different elements:

TiliK225 [7]

3 0

3 years ago

A wave travels at a constant speed. How does the frequency change if the wavelength is reduced by a factor of 4

sweet-ann [11.9K]

The product of (wavelength) x (frequency) is always the same number ... the wave's speed.

So if the wavelength is somehow reduced to 1/4 its original length, the <em>frequency is immediately multiplied by 4</em> . That's the only way their product can remain the same.

4 0

3 years ago

Read 2 more answers

Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at

Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)

C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L

d = 4√2 units of length (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces Fq₁₂ and

Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

In new conditions

Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and

Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx

and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

5 0

3 years ago