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butalik [34]
3 years ago
15

An austrain who lived in vienna who composed graetest music for waltz dance​

Physics
1 answer:
Stolb23 [73]3 years ago
6 0
Johann Strauss II



hope this helps
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Steelhead trout migrate upriver to spawn. Occasionally they need to leap up small waterfalls to continue their journey. Fortunat
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s = 3.26 m

Explanation:

Given that,

Water leaves at a speed of 8 m/s

We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.

At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.

v^2-u^2=2as

a = -g

-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m

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Amoeba

Explanation:A type of protozoan that does not have a definite shape. It moves by using pseudo pods (false feet).

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3 years ago
A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160
Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

P_{f}V_{f}=P_{i}V_{i}

V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}

Put the value into the fomrula

V_{f}=(\dfrac{400}{160})V_{i}

V_{f}=2.5 V_{i}

We need to calculate the work done

Using equation of energy

dQ=dW

dQ=nRTln(\dfrac{P_{in}}{P_{f}})

dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})

dQ=1143\ J

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}

T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}

Put the value into the formula

T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300

T_{f}=207.9\ K

We need to calculate the wok done in adiabatic

Using formula of work done

W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}

W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}

W=574\ J

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

6 0
3 years ago
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