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1 year ago

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1) A person having a mass of 59.1 kg stands before a flight of 30 stairs each of which is 25.0 cm high. He runs up 20 stairs, tu

rns around, walks down 10 stairs, changes his mind and goes up the remaining 20. How much work did he do overcoming gravity? Don’t forget to change the cm to meters.

1 answer:

poizon [28]1 year ago

3 0

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5= 7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= $\frac{W}{T}$

P = 4432.5 /30

P = 147,75 W

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A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m

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Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

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3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

Anna71 [15]

Answer:

B. $V_{f}= 10\,cubic\,inches$

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

$PV=nRT$

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

$P_{i}V_{i}=n_{i}RT_{i}$ (1)

$P_{f}V_{f}=n_{f}RT_{f}$ (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because $T_{i}=T_{f}$ , $n_{i}=n_{f}$ and R is an universal constant:

$P_{i}V_{i}= P_{f}V_{f}$ , solving for $V_{f}$

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3 years ago

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2 years ago

A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

Sonbull [250]

Answer:

(a) $\overrightarrow{L}=885.5\widehat{k}$

(b) $\overrightarrow{L}=1046.5\widehat{k}$

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, $\overrightarrow{r}=3\widehat{i}-4\widehat{j}$

$\overrightarrow{p}=m\overrightarrow{v}$

$\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )$

$\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}$

So, the angular momentum

$\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=885.5\widehat{k}$

(b) Here, $\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}$

$\overrightarrow{r}=5\widehat{i}-2\widehat{j}$

$\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=1046.5\widehat{k}$

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3 years ago