All categories

3 years ago

What should he do next in order to follow the steps of inquiry?

Physics

1 answer:

andrezito [222]3 years ago

6 0

Since he wonders, then according to the formal, methodical procedures of scientific inquiry, he should not go running off and trying stuff before he <em>forms a hypothesis that can be tested.</em>

In his mind, at some point, he should say to himself <em>"Self ! It seems to me that if salt is added to water, it makes the water boil at a higher temperature than pure water does."</em>

And only THEN, with this statement in mind, he's ready to design an experiment to test it.

You might be interested in

What is the name for family labeled #4 (Yellow)?

goldfiish [28.3K]

Answer:

transition metals im sorry if this was too late

4 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

Arisa [49]

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

= 1/2 (6 X 15)

= 1/2 (90)

= 45 meters

Hence, the object traveled 45 meters.

4 0

4 years ago

There are two nearby point sources, each emitting a light wave of frequency f. When the frequency f is increased, how will the d

Degger [83]

Answer:

An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough

Explanation:

This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.

6 0

3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4

Dima020 [189]

The rate of energy radiated by the man is 3.86 x $10^{-8}$ J/s. $m^{2}$ .

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ $T^{4}$

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

$\frac{Q(t)}{T^{4} }$ = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x $10^{-8}$ J/s.

So that;

$\frac{Q(t)}{T^{4} }$ = 1.7 * 0.4 * 5.67 x $10^{-8}$

= 3.8556 x $10^{-8}$

= 3.86 x $10^{-8}$ J/s. $m^{2}$

Thus, the rate of energy radiated by the man is 3.86 x $10^{-8}$ J/s. $m^{2}$ .

Learn more on energy radiation of objects by visiting: brainly.com/question/12550129

4 0

3 years ago