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vlabodo [156]
2 years ago
9

Hope you can answer this: A Student Visits A Farm And Makes These Notes In Her Journal.

Physics
1 answer:
Strike441 [17]2 years ago
8 0

Answer:

Common Sense

Explanation:

The chick has probably seen other chicks get caught by the Hawk and knows not to go near, or saw a giant bird flying straight towards it and used common sense to identify it as danger and run away. Although if this is for a test or a grade or something, please do not use the answer, it is most likely incorrect. This is honestly my best answer.

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A skydiver finds that she speeds up when she holds her arms close to her body. What does this do?
ArbitrLikvidat [17]

D. Reduces the force of air resistance

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2 years ago
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A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting
Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

\Delta U = n C_v(T_f-T_i)

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2    [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K

So,  \Delta U= 0.2\times12.5(1200-300)\\=2250 J

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J

Since, Q is positive 2250 J of heat will flow into the system.

6 0
2 years ago
Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is
pantera1 [17]

the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

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2 years ago
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a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient
Rufina [12.5K]

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵   (ºC)⁻¹

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sergey [27]

Answer:0.502kg

Explanation:

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1260 * 15 *60 = M * 22.6 * 10^5

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3 0
2 years ago
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