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Elenna [48]
3 years ago
11

2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did

the person deliver?​
Physics
1 answer:
Aneli [31]3 years ago
4 0

Answer:

<em>The person delivered a power of 2,558 Watt</em>

Explanation:

<u>Work and Power</u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being  the force vector and  the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

\displaystyle P=\frac {30,184}{11.8}

P = 2,558 Watt

The person delivered a power of 2,558 Watt

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The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How f
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Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

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3 years ago
Two moles of helium gas initially at 438 K and 0.44 atm are compressed isothermally to 1.61 atm. Find the final volume of the ga
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44.64335 L

Explanation:

R = Gas constant = 8.314 J/mol K = 0.08205 L atm/mol K

P = Pressure

V = Volume

T = Temperature = 438 K

1 denotes initial

2 denotes final

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2\times 0.08205\times 438}{0.44}\\\Rightarrow V=163.35409\ L

So,

P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{0.44\times 163.35409}{1.61}\\\Rightarrow V_2=44.64335\ L

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3 years ago
What is the rate at which an object moves but doesn't include direction?
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about the direction it's moving, is the object's SPEED .

When the direction is also given, then you have the object's VELOCITY.


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6 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
2 years ago
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