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2 years ago

Observations must be _____. subjective objective biased deductive

Physics

2 answers:

Dmitrij [34]2 years ago

6 0

Objective
because it means<span> based on measurement, or reasoning free of bias.</span>

aivan3 [116]2 years ago

3 0

Answer: Objective

An observation is the outcome of the experimentation procedure. It is based upon the measurements and virtual evidences obtained while applying procedures. The observations should be objective that means they should be noted or recorded as such without any manipulation. An analyst should not make assumptions and the observation should be unbiased for a particular parameter because results will not be accurate.

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W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively

MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

ΔU = ΔK

7 0

3 years ago

How do you draw an ear

Neporo4naja [7]

Answer:

drawing of ear step by step

8 0

3 years ago

A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from

Naddik [55]

Complete Question

The complete question iws shown on the first uploaded image

Answer:

$y \to z \to x$

$x \to z \to y$

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

Now the magnetic force is mathematically represented as

$F = I L$ x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

The motion of the loop is from y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

The motion of the induced current will now be x to z to y

5 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

2 years ago