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WINSTONCH [101]
3 years ago
11

State pascal is law.please answer me​

Physics
1 answer:
Angelina_Jolie [31]3 years ago
5 0

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions

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Cual es la fuerza de empuje que recibe un cuerpo de forma cilíndrica que mide 0.3 m de radio y 8m de altura, el cual se sumerge
zimovet [89]

Responder:

Fb = 17,5 kN

Explicación:

La expresión de la fuerza de empuje

Fb = ρgV ---------- 1

dónde

Fb = fuerza de flotación

ρ = densidad del fluido

g = aceleración debida a la gravedad

V = volumen de fluido

Paso uno:

datos dados

radio del cilindro = 0,3 m

altura h = 8m

densidad = 790 kg / m ^ 3

g = 9,8 m / s ^ 2

El volumen del cilindro se expresa como

V = πr ^ 2h

V = 3,142 * 0,3 ^ 2 * 8

V = 3,142 * 0,09 * 8

V = 2,26

El volumen es 2.26m ^ 3, el volumen del cilindro es igual al volumen de alcohol disparado

La expresión de la fuerza de empuje es

Fb = ρgV

sustituyendo tenemos

Fb = 790 * 9,8 * 2,26

Fb = 17514,26N

Fb = 17,5 kN

3 0
3 years ago
How long would a simple pendulum need to be to have a period of 1. 0s.
Archy [21]

Answer:

0.248 m

Explanation:

5 0
2 years ago
How are reversibility, time, and entropy connected
kondor19780726 [428]

Answer:

As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease. ... Thus, entropy measurement is a way of distinguishing the past from the future.

Explanation:

6 0
3 years ago
I’m not accelerating, so the net (vertical) force on me, while I’m sitting here doing this lab is _________
Elina [12.6K]

Answer:

Net force is zero

Explanation:

According to the Newton's second law, the net force on the body is equal to the product of mas of body and the acceleration.

here acceleration is equal to zero so net force is also zero because mass of an object can never be zero.

3 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
4 years ago
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