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3 years ago

28. Identify whether the following objects are in

Physics

1 answer:

Marianna [84]3 years ago

7 0

Answer:

a. A baseball after it has been hit - not in free fall

b. A rock that is thrown in the air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
If the body is under free-fall, but a centripetal force acts on it where it is equal to the gravitational force at that point. The object will have two components of velocity along the tangential line, perpendicular to the radius of the orbit.

a. A baseball after it has been hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

d. A paper airplane - not in free fall according to point 1 & 2.

e. A bird flying - not in free fall according to point 1 & 2.

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Answer:

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Explanation:

From the question we are told that

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The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

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Explanation:

Given that,

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$W=\dfrac{1}{2}m(v^2-u^2)=0$

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

$W=Fd\ cos\theta$

Here, $\theta=90+8=98^{\circ}$

$W=mgd\ cos\theta$

$W=60\times 9.8\times 300\ cos(98)$

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

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$W_f=-W$

$W_f=24550.13\ J$

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