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3 years ago

Geophysicists have estimated that the temperature at the center of the Earth's

Physics

1 answer:

dem82 [27]3 years ago

3 0

Answer: B. 9032°F

Explanation:

First we take down the data made available to us.

Estimated temperature of the Earth's core = 5000° Celsius

Now to express this temperature in Fahrenheit, we make use of the formula;

[°F] = [°C] × 9/5 + 32

Where

[°F] is temperature in Fahrenheit and

[°C] is temperature in Celsius

So we just input our figure into the formula

[°F] = [5000°] × 9/5 + 32

°F = 5000 × 1.8 + 32

°F = 9000 + 32

°F = 9032°

So 5000° Celsius is expressed as 9032° Fahrenheit

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A standard baseball has a circumference of apoximately 23cm. If a baseball had the same mass per unit volume as a neutron or a p

BigorU [14]

Baseball, neutrons, and protons are all perfect spheres.

Circumference of the baseball, $L_{\text {ball }}=23 \mathrm{~cm}$ .

mass of the proton or neutron, $m_{\mathrm{p}}=10^{-27} \mathrm{~kg}$

diameter of the proton or neutron, $r_{0}=10^{-15} \mathrm{~m}$

Mass per unit volume is the definition of a material's density. The density equation is

$\rho=\frac{M}{V} .$

Here, M is the mass of the material, V is the volume, and $\rho$ is the density of the material.

We know that formula for the volume of a sphere having diameter d is $V=\frac{\pi d^{3}}{6}$.$

You may obtain this result by plugging this number into the proton/neutron density equation:

$\rho_{0}=\frac{m_{0}}{\left(\frac{\pi d_{0}^{3}}{6}\right)}$

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}\rho_{\circ} &=\frac{10^{-27}}{\left(\frac{\pi\left(10^{-15}\right)^{3}}{6}\right)} \\&=1.91 \times 10^{18} \mathrm{kgm}^{-3}\end{aligned}$

The formula for calculating a circle's circumference with diameter d is $L=\pi d$.$

Using this, you get the diameter of the baseball:

$\begin{aligned}d_{\text {ball }} &=\frac{L_{\text {ball }}}{\pi} \\&=\frac{23}{\pi} \\&=7.32 \mathrm{~cm} \\&=0.0732 \mathrm{~m}\end{aligned}$

The volume of the baseball may be calculated using the formula for the volume of a sphere:

$\begin{aligned}V_{\text {ball }} &=\frac{\pi d_{\text {ball }}^{3}}{6} \\&=\frac{\pi \times(0.0732)^{3}}{6} \\&=2.054 \times 10^{-4} \mathrm{~m}^{3}\end{aligned}$

The mass of the baseball is determined by the concept of density if you suppose that its density is equal to that of a proton or neutron.

$m_{\text {ball }}=\rho_{\circ} \cdot V_{\text {ball }}$

Here, $\rho_{\circ}$ is the density of the proton/neutron.

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}m_{\text {ball }} &=\left(1.91 \times 10^{18}\right) \cdot\left(2.054 \times 10^{-4}\right) \\&=3.92 \times 10^{14} \mathrm{~kg}\end{aligned}$

Therefore, if the density of the baseball was equal to that of the neutron/proton, then its mass would be $3.92 \times 10^{14} \mathrm{~kg}$.$

Learn more about density brainly.com/question/15164682

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7 0

1 year ago

A race cars velocity increases from 6.0 m/s to 36 m/s over a 5.0 s time interval. What is It’s average acceleration?

Sonja [21]

The acceleration of the race car is 6 m/s²

The given parameters;

initial velocity of the race car, u = 6 m/s

final velocity of the race car, v = 36 m/s

time of motion, t = 5 s

Acceleration is defined as the change in velocity per change in time of motion.

The acceleration of the race car is calculated as follows;

$a=\frac{v-u}{t} \\\\a = \frac{36-6}{5} \\\\a = 6 \ m/s^2$

Thus, the acceleration of the race car is 6 m/s²

Learn more here: brainly.com/question/17280180

3 0

3 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0

3 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago

visible light is a range of ____ energy EM waves in the electromagnetic spectrum that the human eye can see

Artemon [7]

A range of infrared wavelengths <span />

4 0

3 years ago