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3 years ago

Can the big bang's sound still be heard?

1 answer:

4 0

There is no sound in space because there is no medium to move it. However i think the big bang can still been seen in the form of background radiation but i am unsure.

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The longest banana split ever made was 7.32 km long (obviously they used more than one banana). If an archer were to shoot an ar

elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

8 0

3 years ago

A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65

Sedaia [141]

The thermal energy was produced is 116J

<h3>What is the thermal energy produced?</h3>

Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;

Ff= μmgcosθ

Ff = 0.65 * 5.0 kg * 9.8 m/s^2 * cos 25

Ff = 29 N

Hence, the thermal energy is;

29 N * 4.0 m = 116J

Learn more about frictional force:brainly.com/question/1714663

#SPJ1

4 0

2 years ago

Sara went 25 m in 4 s. How fast did she go?

Angelina_Jolie [31]

She went around half a microsecond

8 0

3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago

22. When looking at the halogens on the periodic table, how does the number of valence electrons change as the

Gennadij [26K]

Answer:

The total number of valence electrons per atom does not change. It remains at 7 (as opposed to 1 in the options.)

Explanation:

The periodic table organizes elements with the same number of valence electrons in the same column.

All the halogens (F, Cl, Br, etc.) are found in the same column (the second column from the right) on the periodic table. The all have the same number of valence electrons per atom.

All elements in the right-most column (the one with all the noble gas elements) contain 8 valence electrons per atom (a full valence shell.) Since the halogens are to the left of the noble gas elements by only one place, each would contain $(8 - 1) = 7$ valence electrons.

7 0

3 years ago