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3 years ago

7

Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea

rth. What will happen to the weight of the telescope when it reaches the moon? The weight of the telescope will increase. The weight of the telescope will decrease. The weight of the telescope will stay the same. The weight of the telescope will be equal to its mass.

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

Explanation:

the weight of the telescope decreases because the moon attract the body with less force as compared to earth due to less gravity as compared to earth

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Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

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$A_{o}$ is the initial amount of the material

$t$ is the time elapsed

$h=138 days$ is the half life of polonium-210

Knowing this, let's substitute the values and find $t$ from (1):

$\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}}$ (2)

$\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}}$ (3)

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$-1.386=-\frac{t}{138days}ln(2)$ (6)

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$t=276days$ (7)

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A 1-m3 tank containing air @ 25 oC & 500 kPa is connected to another tank containing 5 kg of air at 35 oC & 200 kPa thro

VladimirAG [237]

Answer:

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Equilibrium Pressure, $P_{eq} = 278.82 kPa$

Given:

Volume of Tank 1, V = 1 $m^{3}$

Temperature of Tank 1, T = $25^{\circ}C$ = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = $35^{\circ}C$ = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, $20^{\circ}C$ = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

$m' = \frac{PV}{RT}$

$m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg$

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = $\frac{mRT}{P}$

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Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17 $m^{3}$

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

$P_{eq}V'' = m''RT_{eq}$

$P_{eq} = \frac{m''RT_{eq}}{V''}$

$P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa$

$P_{eq} = 287.82 kPa$

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