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3 years ago

12

Can you please help me with #9?

1 answer:

IRINA_888 [86]3 years ago

3 0

The answer is 1 m/s. :::))))

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Message body weight 30 points

faust18 [17]

one hundred and fourty-six pounds

4 0

3 years ago

A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state. T

Flauer [41]

Answer:

C. identity of the element

The specific amount of energy emitted when electrons jump from excited states to the ground state refers to emission spectrum. The energy is emitted in the form of photons, and the photons have very specific wavelengths (energy) that correspond to the energy gaps between the excited states and the ground state. The specific wavelengths of light emitted are referred to as the "emission spectrum," and each element produces a different emission spectrum. Thus, this emitted energy can be used to identify the element from which your sample was taken.

Explanation:

6 0

1 year ago

What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9

mr_godi [17]

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

$v_f^2 = v_i^2 + 2 a \Delta y$

where

$v_f$ is the final velocity

$v_i$ is the initial velocity

a is the acceleration

$\Delta y$ is the distance covered

For the particle in free-fall in this problem, we have

$v_i = 0$ (it starts from rest)

$v_f = 7.9 m/s$

$g=9.8 m/s^2$ (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

$\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m$

5 0

3 years ago

A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?

ivolga24 [154]

Answer:

Explanation:

Given

mass of rock $m=40\ kg$

Elevation of Rock $h=10\ m$

Distance traveled by rock with time

$h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

$5=0\times t+\frac{1}{2}(9.8)(t^2)$

$t^2=\frac{10}{9.8}$

$t=1.004\approx 1\ s$

velocity at this time

$v=u+at$

$v=0+9.8\times 1.004$

$v=9.83\ m/s$

6 0

3 years ago

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

$A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})$

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

$A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )$

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

7 0

3 years ago