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2 years ago

Pls help me l will make it brainlest

Physics

2 answers:

emmasim [6.3K]1 year ago

7 0

Calculate the value of the area expansion of a steel rod of lenght 4cm at 20°c after heating attains another length of 9cm at 70°c.

Area = 0.02

Area Expansity = $0.0 {4}^{ - 1}$

nikklg [1K]1 year ago

3 0

Answer:

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

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=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = <u>0.02</u>

============================================================

Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

⇒ β = 2 × α

⇒ β = 2 × 0.02

⇒ β = <u>0.04 °C⁻¹</u>

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³

Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

= 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

9.5 N = 10 N - weight of water displaced

Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

= 0.051 kg

Therefore the volume of the water displaced,

V = m / ρ

= 0.051 / 1000

= 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

= 5.1 x 10⁻⁵ x 800

= 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

= 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

W₀ = W - w

= 10 N - 0.4 N

= 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0

3 years ago

How much would you have to eat for your appendix to explode ? I ate so much on thanksgiving and i need to know i’m very scared

dimulka [17.4K]

Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

6 0

3 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

Compare physical changes and chemical changes

JulijaS [17]

Answer:

The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to physical "reactions" as physical changes from now on.

Explanation:

Physical changes are limited to changes that result in a difference in display without changing the composition. Some common changes (but not limited to) are:

Texture

Color

Temperature

Shape

Change of State (Boiling Point and Melting Point are significant factors in determining this change.)

Physical properties include many other aspects of a substance. The following are (but not limited to) physical properties.

Luster

Malleability

Ability to be drawn into a thin wire

Density

Viscosity

Solubility

Mass

Volume

4 0

3 years ago

Which force is always pulling but never pushes and is directly influenced by mass

Lapatulllka [165]

<h2>Answer: I know when it comes to magnetic objects the magnet always pulls not push.</h2>

7 0

3 years ago

Read 2 more answers

Pls help me l will make it brainlest ​

Pls help me l will make it brainlest