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3 years ago

What was the first logical for the ways things moves in the solar system?

Physics

1 answer:

inn [45]3 years ago

6 0

Answer:

depends the ancient humans or the start of the solar system

Explanation:

if its the humans the earth was in the middle of everything.

but the beginning of the solar system lets say when earth just formed the orbits where unstable and planet ending impacts happed a lot.

ill do the proto planetary disc too

the disc formed the gas giants first then the rockey planets formed out of the disc

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For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg

jeyben [28]

Answer:

<h2>Magnetic field strength in that region is 1.2 T</h2>

Explanation:

As we know by the formula of radius of charge moving in external field is given as

$R = \frac{mv}{qB}$

so we will have

$R = 25 cm$

$m = 1.6 \times 10^{-27} kg$

$q = 1.6 \times 10^{-19} C$

$v = 0.10 \times 3 \times 10^8 m/s$

now we have

$0.25 = \frac{(1.6 \times 10^{-27})(3\times 10^7)}{(1.6 \times 10^{-19})B}$

now we have

$B = 1.2 T$

8 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago

A flea jumps by exerting a force of 1.09 ✕ 10−5 N straight down on the ground. A breeze blowing on the flea parallel to the grou

patriot [66]

Answer:

a= 17.877 m/s² : Magnitude of the acceleration of the flea

β = 88.21° : Direction of the acceleration of the flea

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

Look at the flea free body diagram in the attached graphic

The acceleration is presented in the direction of the resultant force (R) applied over the flea .

$R= \sqrt{(F_{x})^{2} + (F_{y})^{2} }$

$R= \sqrt{(10.9)^{2}+(0.340)^{2} } *10^{-6} N$

R= 10.905*10⁻⁶ N

We apply the formula (1) to calculate the magnitude of the acceleration of the flea

∑F = m*a m = 6.1 * 10⁻⁷ kg

R = m*a

a= R/m

a= (10.905*10⁻⁶) / (6.1 * 10⁻⁷ )

a= 17.877 m/s²

β: Direction and magnitude of the acceleration of the flea

$\beta = tan^{-1} (\frac{F_{y} }{F_{x} } )$

$\beta = tan^{-1} (\frac{10.9*10^{-6} }{0.340*10^{-6} } )$

β = 88.21°

5 0

4 years ago

A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at

4vir4ik [10]

Answer:

0.541 nm

Explanation:

The condition for maxima is,

$dsin\theta=m\lambda$

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, $\lambda$ is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

$\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m$

And the order of maxima is $m=1$ .

And the angle at which first order maxima occur is, $\theta=33^{\circ}$ .

Put these values in maxima condition while solving for d.

$d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm$

Therefore, the slit separation is 0.541 nm.

7 0

3 years ago

How do we see???????

SVETLANKA909090 [29]

Our eyes.... they send in image to our brain that's how we see... this isn't sarcasm

7 0

3 years ago

Read 2 more answers