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viva [34]
3 years ago
11

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

= 239 m/s. Upon landing, the plane can produce an average deceleration of a = 16.5 m/s².
How long will it take the plane to circle the Earth at the equator?
Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

R= 6370*10^3 m

v = 239m/s

a = 16.5m/s^2

The circumference of the earth would be

\phi = 2\pi R

Velocity is defined as,

v = \frac{x}{t}

t = \frac{x}{v}

Here x = \phi, then

t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}

t = 167463.97s

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

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You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the res
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Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55

Add them all together to get the x component of the resultant vector, V:

V_x=55

Do the same to find the y components of the part of this journey:

A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12

Add them together to get the y component of the resultant vector, V:

V_y=88

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

V_{mag}=\sqrt{55^2+88^2} and, rounding to 2 sig dig's as needed:

V_{mag}= 1.0 × 10² m; now for the direction:

\theta=tan^{-1}(\frac{88}{55})= 58°

7 0
3 years ago
What electric force would a stationary 3.8 C charge experience if it were far away from any other charges
MAVERICK [17]

Answer:

The electric force will be  0 N

Explanation:

From the question we are told that

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Generally from Coulombs law the electric force  between two charges is mathematically represented as

         F = \frac{ k  *  q_1 q_2 }{r^2}

Here r is the distance of separation between that two charges.

  Now from the question we are told that the charge is far away from any other charge hence we can say that the distance between the charge and any other charge is  r = \infty

So

       F = \frac{ k  *  3.8  * q_2 }{\infty^2}

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Hence the electric force will be  0 N

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Answer:

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Explanation:

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PIT_PIT [208]
The object will continue moving in a straight line at constant speed.
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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
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Explanation:

We have,

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