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3 years ago

Two forces are acting on an object, but the net force on the object is O N. For the net

Physics

2 answers:

Alenkinab [10]3 years ago

6 0

Answer:

A. The forces are the same size and in opposite directions.

Explanation:

Just as an opposite number will cancel a number: -1 +1 = 0, so an opposite force will cancel a force, with the result that the net is zero.

maria [59]3 years ago

4 0

Its A The force are the same size and in opposite direction

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Be sure to answer all parts. Consider the following energy levels of a hypothetical atom: E4 −2.01 × 10−19 J E3 −5.71 × 10−19 J

pashok25 [27]

(a) 159 nm

First of all, let's calculate the energy difference between the level E1 and E4:

$\Delta E=E_4 -E_1 = -2.01\cdot 10^{-19}J-(-1.45\cdot 10^{-18} J)=1.25\cdot 10^{-18} J$

Now we know that this energy difference is related to the wavelength of the absorbed photon by

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

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the photon

Solving for $\lambda$ , we find

$\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.25\cdot 10^{-18} J}=1.59\cdot 10^{-7} m = 159 nm$

b) 293 nm

As done in part a), let's calculate the energy difference between the level E2 and E3:

$\Delta E=E_3 -E_2 = -5.71\cdot 10^{-19}J-(-1.25\cdot 10^{-18} J)=6.79\cdot 10^{-19} J$

this energy difference is related to the wavelength of the absorbed photon by

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

Solving for $\lambda$ again, we find

$\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{6.79\cdot 10^{-18} J}=2.93\cdot 10^{-7} m = 293 nm$

c) 226 nm

As done in part a) and b), let's calculate the energy difference between the level E1 and E3:

$\Delta E=E_3 -E_1 = -5.71\cdot 10^{-19}J-(-1.45\cdot 10^{-18} J)=8.79\cdot 10^{-19} J$

this energy difference is related to the wavelength of the emitted photon by

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

Solving for $\lambda$ again, we find

$\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{8.79\cdot 10^{-18} J}=2.26\cdot 10^{-7} m = 226 nm$

4 0

3 years ago

malfutka [58]

Answer:

The penny will hit the ground at 6.39 seconds

Explanation:

Free Fall

The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:

$\displaystyle y=\frac{gt^2}{2}$

Where $g=9.8\ m/s^2$ , and t is the time.

Solving for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Using the value y=200:

$\displaystyle t=\sqrt{\frac{2*200}{9.8}}$

t=6.39 sec

The penny will hit the ground at 6.39 seconds

4 0

4 years ago

How many electrons does a neutral atom of oxygen-18 have? 8 O 16.00

galina1969 [7]

Answer:

A. 8

Explanation:

Atomic number of oxygen = 8

Oxygen is the 8th element in periodic table.

Atomic no. = 8, means one oxygen atom has 8 protons. Now, because the atom is neutral, so number of electrons = no. of protons in an atom.

Hence, the number of electrons in a neutral oxygen atom is 8.

7 0

3 years ago

Read 2 more answers

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$

where

$\lambda$ is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

$\lambda =2(5.00)=10.0 m$

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$

And by comparing it with the general equation of a wave:

$y(t)=A sin (\frac{2\pi}{T} t)$

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

$sin(\frac{2\pi}{T}t)=1$

which means that

$y(t)=A$

And therefore in this case,

$y=0.500 cm$

So, this is the displacement.

6 0

3 years ago

The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn

valentinak56 [21]

Assume: neglect of the collar dimensions. Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa Ď„=(S*Q)/(I*b)=(40*ă€–10ă€—^3*Ď€(ă€–0.125ă€—^2-ă€–0.117ă€—^2 )*121*ă€–10ă€—^(-3))/(Ď€/2 (ă€–0.125ă€—^4-ă€–0.117ă€—^4 )*8*ă€–10ă€—^(-3) )=41.277 MPa @ Point K: Ď_z=(+M*c)/I=(40*0.6*121*ă€–10ă€—^(-3))/(8.914*ă€–10ă€—^(-5) )=32.6 MPa Using Mohr Circle: Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 ) Ď_max=104.2 MPa, Ď„_max=45.62 MPa

3 0

4 years ago