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2 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]2 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]2 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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A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block $\rho_{b} =9.50\times10^2\ kg/m^3$

Density of fluid $\rho_{t} =1.30\times10^3\ kg/m^3$

We need to calculate the depth

Using balance equation

$mg=\rho g V$ ....(I)

We know that,

The density is

$\rho=\dfrac{m}{V}$

$m=\rh0\times V$

Put the value of m in equation (I)

$\rho_{b}\times V\times g=\rho_{t}\times g\times V$

$\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h$

$h=\dfrac{\rho_{b}H}{\rho_{t}}$

Put the value into the formula

$h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}$

$h= 5.85\ cm$

$h=0.1919291\ ft$

We need to calculate the acceleration

Using formula of net force

$F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g$ ....(II)

Put the value in the equation (II)

$a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8$

$a=3.61\ m/s^2$

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

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