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3 years ago

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45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d

isplacement at t = 5 s? (b) What is its velocity at this same time?

2 answers:

omeli [17]3 years ago

7 0

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

Brut [27]3 years ago

7 0

Answer:

Velocity at 5 seconds: 180 m/s

Ditance: 525 meters

Explanation:

In order to calculate displacement in this option we need to use the formula for distance when an object is experimenting aceleration:

$s=VoT+1/2at^2$

So now we just have to insert the data that we know:

$s=VoT+1/2at^2\\s= 30*5+ 1/2*30*25\\s=150+15*25\\s= 150+375\\s=525 m$

The distance after 5 seconds with an initial velocity of 30 m/s and an acceleration of 30 m/s will be 525 meters.

And the final velocity is calculated by the next formula:

$Vf= Vo+ A*t\\Vf= 30 m/s+ 30m/s*5\\Vf=180m/s$

So the final velocity at 5s would be 180 m/s

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asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

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(a) The condition for maxima is given by :

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For third maxima,

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(b) For second dark fringe, n = 2

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3 years ago

A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it

Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given: </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.

<u>Solution </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next

Ef=Ei+W (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form

(Kf + mc^2) = (KJ+ mc^2) (2)

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by

W = FΔr (4)

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut

W = F Δr= (190N)(6 m) = 1140 J

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

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vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W' (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate

1/2mvf'^2=1/2mvi'^2+W'

vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

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Answer:

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Explanation:

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