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3 years ago

The __________ is the layer of the atmosphere where satellites and the space shuttle orbit the Earth.

Physics

2 answers:

Y_Kistochka [10]3 years ago

7 0

Answer: d. exosphere

The exosphere is the outermost layer of the atmosphere of earth. It is present above the thermosphere which is lying above the earth. From this layer, the atoms and molecules escape out, which maintains an environment for the movement of satellites and space shuttles.

Gemiola [76]3 years ago

3 0

The most likely answer, according to an informational article I have read, on the earths atmosphere, is exosphere. Hope it helps! If u need more info, message me!

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A 900-N lawn roller is to be pulled over a 5-cm (high) curb. Radius of roller is 25 cm. What minimum pulling force is needed if

Feliz [49]

I believe b 30 degrees

8 0

3 years ago

A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –

zhenek [66]

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

$W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J$

So, the work done by the given force is 18.88 J.

5 0

3 years ago

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

$E=\frac{hc}{\lambda}$

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

5 0

3 years ago

Question 5 of 10

muminat

The correct answer to this question is D

5 0

3 years ago

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

3 years ago