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CaHeK987 [17]
3 years ago
11

When does an object falling vertically through the air reach terminal velocity?

Physics
1 answer:
zlopas [31]3 years ago
7 0

Answer:

a. when the acceleration of the objects become negative

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A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 3
ryzh [129]

Answer:

The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is,E=69\ GPa=69\times 10^9\ Pa

The stress acting on material is proportional to the strain. Its formula is given by :

\epsilon=\dfrac{\sigma}{E}

\sigma is the stress

\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}

Thus, The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

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4 years ago
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An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre
irakobra [83]

Answer:

T = -282.33^o C

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)

here we know that

P_1 = 6.50 \times 10^4

P_o = 4.80 \times 10^4

T_1 = 100^o C

T_o = 0.01^o C

now we will have

P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)

P = 4.80 \times 10^4 + 170.02(T - 0.01)

now if P = 0

then we will have

0 = 4.80 \times 10^4 + 170.02(T - 0.01)

T = -282.33^o C

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The north and North Pole or the south and the South Pole are facing each other which is causing them to repel
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True or Flase The fastest moving traffic on the expressway will be traveling in the right lane
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