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3 years ago

It is being thrown up from 0m

Physics

2 answers:

ohaa [14]3 years ago

8 0

Answer:

tair = 4.81 s

vi = 45.6 m/s

θi = 76.8°

Explanation:

In the x direction:

x₀ = 50 m

x = 0 m

v₀ = -vi cos θ

a = 0 m/s²

v = -vf cos 15° = -0.966 vf

Equation 1:

x = x₀ + v₀ t + ½ at²

0 = 50 + (-vi cos θ) t + ½ (0) t²

0 = 50 − vi cos θ t

Equation 2:

v² = v₀² + 2a(x − x₀)

(-0.966 vf)² = (-vi cos θ)² + 2 (0) (0 − 50)

0.966 vf = vi cos θ

vf = 1.035 vi cos θ

In the y direction:

y₀ = 0 m

y = 100 m

v₀ = vi sin θ

a = -9.8 m/s²

v = -vf sin 15° = -0.259 vf

Equation 3:

y = y₀ + v₀ t + ½ at²

100 = 0 + (vi sin θ) t + ½ (-9.8) t²

100 = vi sin θ t − 4.9t²

Equation 4:

v² = v₀² + 2a(y − y₀)

(-0.259 vf)² = (-vi sin θ)² + 2 (-9.8) (100 − 0)

0.067 vf² = vi² sin² θ − 1960

4 equations, 4 variables. Start by rearranging the first and third equations for vi cos θ and vi sin θ:

vi cos θ = 50 / t

vi sin θ = (100 + 4.9t²) / t

Substitute into the second and fourth equations:

vf = 1.035 (50 / t) = 51.76 / t

0.067 vf² = ((100 + 4.9t²) / t)² − 1960

Substitute the expression for vf and solve for t:

0.067 (51.76 / t)² = ((100 + 4.9t²) / t)² − 1960

179.49 / t² = (100 + 4.9t²)² / t² − 1960

179.49 = (100 + 4.9t²)² − 1960t²

179.49 = 10000 + 980t² + 24.01t⁴ − 1960t²

0 = 24.01t⁴ − 980t² + 9820.51

Solve with quadratic formula:

t² = 17.67 or 23.14

t = 4.20 or 4.81

If t = 4.20:

vi cos θ = 11.89

vi sin θ = 44.39

Divide second equation by first:

tan θ = 3.73

θ = 75.0°

vi = 46.0 m/s

Else, if t = 4.81:

vi cos θ = 10.39

vi sin θ = 44.36

Divide:

tan θ = 4.27

θ = 76.8°

vi = 45.6 m/s

Graph:

desmos.com/calculator/yvmr3tkp28

From the graph, we see the first solution is extraneous (the final θ is -15°).

So only the second solution is correct.

Citrus2011 [14]3 years ago

3 0

Explanation:

In the picture above.

I hope that it's a clear solution.

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An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

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Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

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where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

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$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

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replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

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$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

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It is being thrown up from 0m ​

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