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harina [27]
3 years ago
13

An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s

. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?
Physics
1 answer:
guapka [62]3 years ago
6 0

Answer:

a) Magnitude = 1.03 m/s², Direction: south

b) V_{f} = 8.16 m/s

Explanation:

a) The magnitude and direction of the acceleration can be calculated using the following equation:

V_{f} = V_{0} + at     (1)

Where:

V_{f}: is the final speed = 9.40 m/s  

V_{0}: is the initial speed = 13.0 m/s

t: is the time = 3.50 s

Solving equation (1) for a, we have:

a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2}

Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating.  

b) The final velocity of the bird can be found using the same equation 1:

V_{f} = V_{0} + at

V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s

Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.

I hope it helps you!

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The statement “When an object is in orbit, it is falling at the same rate at which the Earth is curving” is true. The speed of a satellite orbiting the earth depends only on the mass of the earth and the mass of the satellite.

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A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

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3 0
3 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
3 years ago
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