What pressure, in millimeters of mercury (mm Hg), is equivalent to 2.13 atmospheres

What is the increase in pressure required to decrease volume of mercury by 0.001%

REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0

3 years ago

The SI system uses three base units. Is this true or false?

ozzi

$The\text{ SI }system\text{ uses seven }base\text{ units, hence the statement is false}$

8 0

1 year ago

Read 2 more answers

A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en

posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. Now please see the picture, which shows the relation between the J and other SI units. Please understand that you can construct your 'own' formulas based these units. Now here is the time saver:

Because almost always the right units are given in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

By this, I hope you now understand the implication of using the right standard SI units and how that can help you figure out what formula is needed.

3 0

3 years ago

An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)

Alecsey [184]

"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is

   (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

   Weight = (60 kg) x (9.8 m/s²)

      =    588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is

(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

   (0.04) x (588 Newtons) = 23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.

6 0

3 years ago

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

See full explanation in the picture. Please rate as brainliest

5 0

3 years ago