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zlopas [31]
2 years ago
6

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. What is the magnitude of the centripetal acceleration during the turn?
a. 12 m/s2
b. 9.8 m/s2
c. 0 m/s2
d. 6.9 m/s
e. 8.1 m/s2
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

The magnitude of the centripetal acceleration during the turn is a=12.04\ m/s^2.

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , T=t\times 4=60\ s.

Now , Velocity ,

v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.

Also , we know :

Centripetal acceleration ,

a=\dfrac{v^2}{r}

Putting all values we get :

a=12.04\ m/s^2.

Hence , this is the required solution .

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Answer:

59cm

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4 0
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A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t
nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

W=mg

where m=100.0 kg is the mass of the box, and g=9.8 m/s^2 is the acceleration due to gravity. Substituting in the formula, we find

W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

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3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

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Substituting, we find

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5 0
2 years ago
Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v
DaniilM [7]

Answer: See attachment below

Explanation:

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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

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Part(b):

Also the angular displacement (\Delta \theta) can be written as

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8 0
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The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere. 

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729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0
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