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Semenov [28]
3 years ago
14

g Power is _________________the force required to push something the work done by a system the speed of an object the rate that

the energy of a system is transformed the energy of a system
Physics
1 answer:
Levart [38]3 years ago
4 0

Answer:

the rate that the energy of a system is transformed

Explanation:

We can define energy as the capacity or ability to do work. Power is defined as the rate of doing work or the rate at which energy is transformed. It can also be regarded as the time rate of energy transfer. In older physics literature, power is sometimes referred to as activity.

Power is given by energy/time. Its unit is watt which is defined as joule per second. Another popular unit of power is horsepower. 1 horsepower = 746 watts.

Very large magnitude of power is measured in killowats and megawatts.

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The equation would be set in acceleration form, and after plugging in, it would read: a= 10 m/s-0m/s divided by 3 seconds. It equaling 3.3 bar notation and units measuring m/s squared. i’M pRobAbLy wRoNg
4 0
3 years ago
Match the microscopes with their useable
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3 years ago
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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston
Kamila [148]
From gas laws:
\frac{PV}{T} = Constant

Therefore,
\frac{ P_{1}  V_{1} }{ T_{1} } =  \frac{ P_{2}  V_{2} }{ T_{2} }

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V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
T_{2} =  \frac{ P_{2}  V_{2}  T_{1} }{ P^{1}  V^{1} } =  \frac{850*350*363.15}{1200*85} = 1059.19 K = 786.04 °C
8 0
3 years ago
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A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Montano1993 [528]

Net Force = (mass) x (acceleration)  (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0
3 years ago
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