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svetlana [45]
4 years ago
8

The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a nee

dle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in kPa
Chemistry
1 answer:
PSYCHO15rus [73]4 years ago
6 0

Answer:

Pressure exerted by the needle = 78.037 kPa

Explanation:

Pressure is defined as the force exerted on a surface per unit area. The SI unit of pressure is Pa (N.m⁻²).

The force exerted by the phonograph needle is equal to the gravitational force.

Thus,

<u>F=m×g</u>

Where,

F is the force

m is the mass of the body

g is the acceleration due to gravity

Also, g = 9.81 ms⁻²

Given, mass of the needle = 1.00 g

The conversion of g into kg is shown below:

1 g = 10⁻³ kg

Thus, mass of the needle = 1×10⁻³ kg

Thus, <u>F= 1×10⁻³×9.81 N = 9.81×10⁻³ N</u>

<u>Ares of the circle = πr²</u>

Given : Radius of tip of the needle = 0.200 mm

The conversion of mm into m is shown below:

1 mm = 10⁻³ m

Thus, mass of the needle = 0.2×10⁻³ m

<u>Thus, Area of the tip of the needle = π×(0.2×10⁻³)² m² = 1.2571×10⁻⁷ m²</u>

So, Pressure =F/A

<u>P = (9.81×10⁻³)/1.2571×10⁻⁷ Pa = 7.8037×10⁴ Pa</u>

The conversion of Pa into kPa is shown below:

1 Pa = 10⁻³ kPa

<u>Thus, pressure exerted by the needle = 7.8037×10⁴×10⁻³ kPa = 78.037 kPa</u>

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alexandr402 [8]

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Explanation:

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  • If a pH is a 7 it is neutral.

In the given question, the pH scale measures for bleach is 8 and for sea water it is 13. So, bleach is basic, not neutral and Sea water is basic too instead of acid. So, Bleach and sea water should be identified as bases.

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3 years ago
What would the solubility of the substance likely be if the water was heated to 150 degrees Celsius ?
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4 years ago
(a) Sketch, in a cubic unit cell, a [111] and a [112] lattice direction. (b) Use a trigonometric calculation to determine the an
fgiga [73]

Answer and Explanation:

a) The direction is shown in the cube diagram attached to this solution.

b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,

Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))

For (111) and (112)

Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))

Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))

Cos Φ = 4/√(3×6)

Cos Φ = 4/√18

Φ = cos⁻¹ (4/√18) = 19.56°

c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!

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11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J
Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

1

∘

C

.

In your case, aluminium is said to have a specific heat of

0.90

J

g

∘

C

.

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

1

∘

C

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

1

∘

C

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

1

∘

C

, you'd have to provide it with

1 gram



0.90 J

+

1 gram



0.90 J

+

...

+

1 gram



0.90 J



10 times

=

10

×

0.90 J

However, you don't want to increase the temperature of the sample by

1

∘

C

, you want to increase it by

Δ

T

=

55

∘

C

−

22

∘

C

=

33

∘

C

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

1

∘

C



10

×

0.90 J

+

1

∘

C



10

×

0.90 J

+

...

+

1

∘

C



10

×

0.90 J



33 times

=

33

×

10

×

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

33

∘

C

will be

q

=

10.0

g

⋅

0.90

J

g

∘

C

⋅

33

∘

C

q

=

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

q

=

m

⋅

c

⋅

Δ

T

, where

q

- the amount of heat added / removed

m

- the mass of the substance

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

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