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Olenka [21]
3 years ago
6

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

= 219 m/s. Upon landing, the jet can produce an average deceleration of a=17 m/s^2. How long will it take the plane to circle the Earth at the equator?
Physics
1 answer:
Anna007 [38]3 years ago
3 0

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

R= 6370*10^3 m

v = 219m/s

a = 17m/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}

Herex = \phi, then

t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{219}

t = 1.82*10^5s

Therefore will take 1.82*10^5 s or 506 hours, 19 minutes, 17 seconds

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Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}

Where K, U, W are linear kinetic energy, gravitational potential energy and work, respectively.

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A boulder with a weight of 780 N is resting at the edge of a cliff that rises 123 m above the ground. What is the gravitational
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Potential energy U = mgh

Given h = 123 m,
mg = F = 780 N

Then
U = (123)(780)
= 95940
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Which is a result of using a machine?
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A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its
ELEN [110]

Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²          

where k = 31 N/m  (spring constant)

s = 0.15 m  (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²    

   

where Ek = Ep = 3.4875J

m = 0.08 kg  (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g          

where F = unknown (frictional force)

∪ = 0.4   (coefficient of friction)

m = 0.08 kg   (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a                  

where F = 0.31392   (frictional force)

m = 0.08 kg   (mass of block)

a = unknown  (acceleration)

a = -3.924 m/s²      -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as            

where  v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

3 0
3 years ago
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