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3 years ago

What allows for the investigation of a variable? Can someone help please.

Physics

2 answers:

ladessa [460]3 years ago

5 0

The answer is <span>C: Experiment </span>

nevsk [136]3 years ago

3 0

Answer: Experiment

Explanation: If you want to study a variable you should do small changes and observe how the variable changes.

This process is called an experiment, where the investigator does small changes and tries to understand how these changes affect the variable. Trying to create a model that describes the variable.

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A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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2 years ago

A young girl is on a swing that completes 35.0 cycles in 25 seconds. What are her frequency and period?

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2 years ago

A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant

Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v $_f$ + xv $_{fg$

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v $_f$ + x₂v $_{fg$

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

3 0

2 years ago