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Inessa [10]
3 years ago
5

The maximum pressure most organisms can survive is about 1000 times the atmospheric pressure. Only small, simple organisms such

as bacteria can survive such high pressures. What then is the maximum depth at which these organisms can live under the sea (assuming that the density of seawater is 1025 kg/m3)
Physics
1 answer:
Schach [20]3 years ago
6 0

Answer:

h = 10000 m

Explanation:

The pressure applied at a depth of the liquid is given by:

P =ρgh

where,

P = Maximum Pressure to Survive = (1000)(Atmospheric Pressure)

P = (1000)(101325 Pa) = 1.01 x 10⁸ Pa

ρ = Density of sea water = 1025 kg/m³

g = 9.8 m/s²

h = maximum depth to survive = ?

Therefore,

1.01 x 10⁸ Pa = (1025 kg/m³)(9.8 m/s²)h

h = (1.01 x 10⁸ Pa)/(1025 kg/m³)(9.8 m/s²)

<u>h = 10000 m</u>

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. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
3 years ago
What is meant by "steam distillation"? choose the correct option from below.
Diano4ka-milaya [45]

Answer:

b is the correct option.

Explanation:

Steam distillation is a distillation in which water is added to the mixture being distilled so that water and the organic compounds vaporize, condense and are collected together. This process is applied for the purification of those materials specially the organic materials which are temperature sensitive. The normal boiling method can not be implemented as these materials starts to decompose on high temperatures. This method has got a lot of applications in industries like petrolium refineries and the oil extraction from some plants etc.

7 0
3 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
Help me find the acceleration
ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.

In the <em>timeless</em> suvat equation,

a = \dfrac{v^2 - u^2}{2\; x}

where

  • a is the acceleration of the car;
  • v is the <em>final</em> velocity of the car;
  • u is the <em>initial</em> velocity of the car; and
  • x is the displacement of the car.

Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.

In this question,

  • a is the unknown;
  • v = -10.9 \; \text{m} \cdot \text{s}^{-2};
  • u = -27.7 \; \text{m} \cdot \text{s}^{-2}; and
  • x = - 105 \; \text{m}.

Apply the <em>timeless</em> suvat equation:

a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}.

The value of a is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0
3 years ago
A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux
bixtya [17]

The total flux through the cylinder is zero.


In fact, the electric flux through a surface (for a uniform electric field) is given by:

\Phi = E A \cos \theta

where

E is the intensity of the electric field

A is the surface

\theta is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.


We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because \theta=90^{\circ} and \cos \theta=0).


On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

\Phi_1 = E \pi r^2

where r is the radius, and where we have taken \theta=0^{\circ} since the perpendicular to the surface is parallel to the direction of the electric field, so \cos \theta=1. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so \theta=180^{\circ} and \cos \theta=-1, therefore the flux is

\Phi_2 = -E \pi r^2

And the net flux through the cylinder is

\Phi = \Phi_1 + \Phi_2 = E \pi r^2 - E \pi r^2=0

4 0
3 years ago
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