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2 years ago

Explain why astronomers divide the sun into different layers.

Physics

1 answer:

strojnjashka [21]2 years ago

8 0

Answer:

cause why not.

the layers exist cause theoretically the deeper you go the hotter it gets and different atoms just do stuff

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The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N

Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

$\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\$

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

$\theta$ = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

$60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\$

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

5 0

3 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

$F= kx$

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

$k= \frac{F}{x}$

$k = \frac{mg}{x}$

$k= \frac{77.2*9.8}{0.0637}$

$k = 11876.92N/m$

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

$T = \frac{2\pi}{\omega}$

Re-arrange to find \omega,

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.14}$

$\omega = 2.93rad/s$

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

$\omega^2 = \frac{k}{m}$

$m = \frac{k}{\omega^2}$

$m = \frac{ 11876.92}{2.93}$

$m = 4093.55Kg$

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

$f = \frac{1}{T}$

$f = \frac{1}{2.14}$

$f = 0.4672 Hz$

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

$t_T = 10*T$

$t_T = 10 *2.14s$

$t_T = 21.4s$

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0

3 years ago

A 50.6 g ball of copper has a net charge of 1.6 µc. what fraction of the copper's electrons have been removed? (each copper atom

Fynjy0 [20]

First figure out how many atoms you have with Avogadro's number. Since there are 63.5 grams/mol and you have 50.6 grams, you have (50.6/63.5)6.022E23=4.7986E23 atoms. Since there are 29 protons per atom, there are also 29 electrons per atom, so you should have a total of
29*4.7986E23=1.3916E25 electrons.
Since there is a positive charge you know some of these electrons are missing. How many are missing can be found by dividing the charge you have by the charge on the electron: 1.6E-6/1.6022E-19 = 9.98627E12 electrons are missing.
Now take the ratio of what is missing to what there should be:
9.98627E12/1.3916E25 = 7.1760873E-13

5 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

A watt is a measure of power (the rate of energy change) equal to 1 j>s. (a) calculate the number of joules in a kilowatt- ho

GalinKa [24]

Before solving this, you must know the definition of these units of measurement. Watt is a measurement of power which is the amount of energy per unit time in seconds. Energy, on the other hand, is expressed through the SI unit Joules. Thus, power is the amount of energy in Joules per second.

From here, you can use the dimensional analysis technique. Also, you should know that 1 kilowatt is equal to 1000 watts, and 24 hours is equal to 86,400 seconds. Then,

100 = Energy/86,400
Energy = 8,640,000 Joules

4 0

3 years ago