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aev [14]
3 years ago
11

Ability of the muscles to function effectively and efficiently without undue fatigue​

Physics
2 answers:
andre [41]3 years ago
8 0

Answer:

Physical fitness can be defined as the “ability to carry out daily tasks with vigor and alertness, without undue fatigue, and with ample energy to enjoy leisure-time pursuits and to meet unforeseen emergencies” (Park 1989).

Explanation:

anastassius [24]3 years ago
3 0

Answer:

Physical fitness can be defined as the “ability to carry out daily tasks with vigor and alertness, without undue fatigue, and with ample energy to enjoy leisure-time pursuits and to meet unforeseen emergencies”

Explanation:

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According to cell theory which of the following is not true?
Anettt [7]

Answer:

first one

Explanation:

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3 years ago
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What are two ways in which the suns energy can be captured and used?
My name is Ann [436]

The oldest way ... the way we've been using as long as we've been
walking on the Earth ... has been to use plants.  Plants sit out in the
sun all day, capturing its energy and using it to make chemical compounds. 
Then we come along, cut the plants down, and eat them.  Our bodies
rip the chemical compounds apart and suck the solar energy out of them,
and then we use the energy to walk around, sing, and play video games.  

Another way to capture the sun's energy is to build a dam across a creek
or a river, so that the water can't flow past it.  You see, it was the sun's
energy that evaporated the water from the ocean and lifted it high into
the sky, giving it a lot of potential energy.  The rain falls on high ground,
up in the mountains, so the water still has most of that potential energy
as it drizzles down the river to the ocean.  If we catch it on its way, we
can use some of that potential energy to turn wheels, grind our grain,
turn our hydroelectric turbines to get electrical energy ... all kinds of jobs. 

A modern, recent new way to capture some of the sun's energy is to use
photovoltaic cells.  Those are the flat blue things that you see on roofs
everywhere.  When the sun shines on them, they convert some of its
energy into electrical energy.  We use some of what they produce, and
we store the rest in giant batteries, to use when the sun is not there.
 
7 0
4 years ago
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A power station with an efficiency e generates W watts of electric power and dissipates D J of heat energy each second to the co
Andrews [41]

Answer: 13.94 tons/s

Explanation:

On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.

Heat energy = mass flow rate * specific heat * Δ T

Q = MC (ΔΦ)

Heat energy, Q= 3.5*10^8J

Mass flow rate, M= ?

Specific heat, C= 4184j/KgC

Change in temperature, ΔΦ= 6°C

M = Q/CΔΦ

M = (3.5*10^8)/4184*6

M = 13942kg/s

M = 13.94 tons/s

3 0
4 years ago
HELP ME PLEASE! according ohm's law if the resistance remains constant what would happen if you decrease the amount of voltage b
SpyIntel [72]
B.
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7 0
3 years ago
Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the
nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

Initial pressure P_{1} = 100 k pa

Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

⇒ Put all the values in the above formula we get the final temperature

⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

6 0
3 years ago
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