You drop a rock off the top of a 100 m tall building. How long does it take to hit the ground?

A delivery drone uniformly slowed down from 45 m/s to 23 m/s in a straight line over a 30-minute duration within a 16-

nikdorinn [45]

Answer: The Answer is A 1.2 x 10−2 m/s2

7 0

3 years ago

An ideal gas in a piston is placed in a refrigerator, which removes 413 J of energy from the gas via cooling. The plunger on the

maria [59]

Answer:

The net change in the internal energy of the gas in the piston is -343J

Explanation:

Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;

∆E = q + w

q = heat transfer

w = workdone

Because heat is lost by the system, the heat transfer is negative

q = -413J

Because work is done on the system, workdone is positive

w = +70J

∆E = -413J + 70J

∆E = -343J

5 0

3 years ago

List out the use of simple machine

jek_recluse [69]

Explanation:

simple machine can multiplayer of speed and force

4 0

2 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

Which correctly lists the outer planets from the least number of moons to the greatest number of moons?

Harlamova29_29 [7]

There are two planets with no moons, Mercury and Venus. After that the order<span> is Earth (1), Mars (2), Neptune (14), Uranus (27), Saturn (62), and Jupiter (67).

so its A

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6 0

4 years ago

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