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Art [367]
2 years ago
7

A 480 g peregrine falcon reaches a speed of 69 m/s in a vertical dive called a stoop. If we assume that the falcon speeds up und

er the influence of gravity only, what is the minimum height of the dive needed to achieve this speed?
Physics
1 answer:
Ivanshal [37]2 years ago
3 0

Answer:

elevation = 243[m]

Explanation:

This problem can be solved using the principle of energy conservation, where potential energy becomes kinetic energy. At the point where the speed is equal to 69 [m / s] will be taken as the reference point for potential energy, at this point all potential energy will have been transformed into kinetic energy.

Ek= 0.5*m*v^{2} \\where:\\m = mass = 480[g] = 0.480[kg]\\v = velocity = 69 [m/s]\\Ek=0.5*.480*(69)^{2} \\Ek=1142.64[J]\\

So this kinetic energy will be equal to the potential energy at the beginning.

Ek=Ep\\Ep=m*g*h\\1142.64=0.480*9.81*h\\h=242.66[m]

The falcon was originally at 243 [m] above the point (reference point) where he reached the speed given.

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In some amazing situations, people have survived falling large distances when the surface they land on is soft enough. During a
OleMash [197]

Answer:

-3413 ft/s2

Explanation:

We need to know the velocity with which he landed on the snow.

He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.

We use the equation of mition

v^2 = u^2 + 2as

v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values

v^2 = 0^2 + 2\times32\times512

v = \sqrt{2\times32\times512} = 128\sqrt{2}

This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.

In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time, u=128\sqrt{2}, v = 0 and s = 4.8 ft.

0^2 = (128\sqrt{2})^2 + 2a\times4.8

a = -\dfrac{2\times128^2}{2\times4.8} = -3413

Note that this is negative because it was a deceleration, that is, his velocity was decreasing.

5 0
2 years ago
A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around
jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground=\mu_s=0.8

\mu_k=0.5

Static friction force,f=\mu N

Normal force=N=mg

Static friction force,f=\mu_s mg=0.8\times 100\times 9.8=784 N

Using g=9.8m/s^2

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0
2 years ago
A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
2 years ago
A starship passes Earth at 80% of the speed of light and sends a drone ship forward at half the speed of light rela- tive to its
alisha [4.7K]

Answer:

Check Explanation

Explanation:

Let the speed of the drone relative to earth be u = ?

Let the speed of the drone relative to the starship be u' = 0.5c

Let the speed of the starship relative to earth be v = 0.8c

In the theory of relativity,

The 3 speeds are related thus

u = (u' + v)/[1 + (u'v/c²)]

u = (0.5c + 0.8c)[1 + ((0.8c × 0.5c)/c²)]

u = 1.3c/[1 + (0.4c²/c²)]

u = 1.3c/1.4 = 0.9286 c = 0.93 c = 93% c

8 0
2 years ago
A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. R
Lady bird [3.3K]

Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The diagram showing the vector has been attached to this response.

As shown in the diagram,

The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e

Dₓ = -D sin θ i   [The negative sign shows that D lies in the negative x direction]

Where;

D = magnitude of D = 155m

θ = direction of D = 23°

Therefore;

Dₓ = -155 sin 23° i

Since Dₓ represents the x component, its unit vector, j component has a value of 0.

Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;

Dₓ = -155 sin 23° i + 0 j

3 0
2 years ago
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