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3 years ago

The fulcrum is the

Physics

1 answer:

olga_2 [115]3 years ago

6 0

Answer:

point of support on which a lever rotates.

Explanation:

The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.

The fulcrum provides the platform for a lever to torque.

The force that opposes motion by the applied force is termed the frictional force.
Friction is a force that opposes motion.
The stored energy of an object is its potential energy.
The potential energy is the energy due to the position of a body.
The distance an object moves when doing work is termed its displacement.

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Which is the best description of biodiversity?

zavuch27 [327]

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2 years ago

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You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the

bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

$PV=nRT$

For Isothermal process:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

$V_{2}=\frac{V_{1}}{1.1}$ (2)

For Isobaric process:

$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}}$ (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

$T_{3}=0.8T_{2}$ (4)

For Isochoric process:

$\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}}$ (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

$P_{4}=1.15P_{3}$ (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

$P_{4}=1.15*1.1P_{1}=1.265P1$

This is, the new pressure of the gas increases by 26.5% with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

$V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1$

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

$T_{4}=1.15*0.8T_{1}=0.92T1$

The new temperature decreases 8% with respect to initial volume.

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3 years ago

Does any one know the answer

valina [46]

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3 years ago

A car goes from 0 to 26.8 m/s in 6.2 s. What is the average acceleration of the car?

Nutka1998 [239]

Answer:

0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.

Explanation:

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3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago