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Dimas [21]
3 years ago
11

• An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is e

xpected that:
(a) the coin will reach the bottom first.
(b) the feather will reach the bottom first.
(c) both objects will reach the bottom at the same instant.

• If this experiment is repeated at a place 2000 kilometers above the sea-level, the acceleration due to gravity gexp is expected to :
(a) increase. (b) decrease. (c) remain constant.
Physics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

c-) both objects will reach the bottom at the same instant.

(b) decrease.

Explanation:

Although the feather is lighter than the coin, the tube where the experiment is performed is evacuated. Therefore there is no air that prevents the feather from falling freely with the same acceleration and speed as the coin.

In fact in the equations of kinematics proposed by Newton, the mass of the bodies is not taken into account, as we can see in the following equation:

v_{f}= v_{i} +g*t

where:

Vf = final velocity [m/s]

Vi = initial velocity [m/s]

g = gravity acceleration [m/s^2]

t = time [s]

Therefore the answer is C.

Gravitational pull is a function of height, as the height of the body increases, the force of gravity decreases.

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Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the
Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0
3 years ago
Match the object with its characteristic.
Triss [41]
What object do you need to match
8 0
3 years ago
Think about holding a glass of cold water. Your hand is warmer than the glass. Do the particles in your hand or those in the gla
choli [55]

technically usually the warmer object/substances particles move master which causes friction among the particles plus the kinetic energy being converted to thermal energy, so i would say the hand.

6 0
3 years ago
A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla
Yuri [45]

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

5 0
3 years ago
How much time will it take a car travelling at 88 km/hr (55mi/hr) to travel 500 km
Illusion [34]
5.6 hours to go 500km
7 0
3 years ago
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