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Tanzania [10]
3 years ago
11

What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?

Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0
Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!
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Position:                 x = 18t    y = 4t - 4.9t²

First derivative:        x' = 18      y' = 4 - 9.8t

Second derivative:    x'' = 0        y'' = - 9.8


Position vector:      P  =  (18t) i  +  (4t - 4.9t²) j

Velocity vector:      V  =  (18) i  +  (4 - 9.8t) j

Acceleration vector  A  =              (- 9.8) j

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

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3 years ago
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AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

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The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

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where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

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