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elena55 [62]
3 years ago
5

A boy throws rocks with an initial velocity of 12m/s [down] from a 20 m bridge into a river. Consider the river to be at a heigh

t of 0 m. Draw a velocity-time and position-time graph. Use up as position and ignore air resistance.
Physics
1 answer:
stira [4]3 years ago
4 0

Answer:

Please find attached the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel

Explanation:

The given parameters are;

The initial velocity with which the boy throws the rock, u = 12 m/s

The direction in which he throws the rock = Down

The height from which the rock was thrown, h = 20 m

The height at which the river is located = 0 m

The kinematic equation of motion, of the rock can be given as follows;

h = u·t + 1/2·g·t²

Where;

t = The time of motion of the rock

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the rock is thrown = 20 m

Substituting the known values into given equation, we get;

20 = 12·t + 1/2·9.8·t² = 12·t + 4.9·t²

4.9·t² + 12·t - 20 = 0

t = (-12 ± √(12² - 4×4.9×(-20)))/(2 × 4.9) = (-12 ± √(242))/(9.8)

t ≈ -3.587 seconds or t ≈ 1.138 seconds

Attached please find the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel.

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Answer:

Creatine

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A man rides a bike along a straight road at a constant speed for 5 minmin, then has a flat tire. He stops for 5 minmin to repair
slamgirl [31]
Use the particle model to drawl a motion diagram
3 0
3 years ago
A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t
azamat

Answer:

(a) x'(t)= 142

(b) 142

(c) y'(t)= -32t+5

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) x(t) = 142t

x'(t)= 142

(b) Since x'(t)= 142 is independent of t, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) y(t) = - 16t^2+5t+5

y'(t)= -32t+5

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to x(t).

x(t)= 142t = 60.5

t=\dfrac{60.5}{142}= 0.426.

In this time, the vertical velocity, y'(t) is

y(t)= -32\times0.426+5 = -8.632

The speed of the ball at thus point is s=\sqrt{142^2+(-8.632)^2}=142 ft/s

To convert this to mph, we multiply the factor 3600/5280

s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}

(e) The time has been determined from (d) above.

t= 0.426

(f) This angle is given by

\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}

\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47 (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}

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CCL4 is Non polar and NH3 is Polar? Why?​
lys-0071 [83]

Answer:

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R
bonufazy [111]

Answer:

Pressure wil be greatest at R [coz due to capillary action, coz it's small cross-sectional area]

6 0
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