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3 years ago

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A 16.75 g piece of iron absorbs 986.75 joules of heat energy, and it’s temperature changes from 25 degrees Celsius to 175 degree

s Celsius. Calculate the specific heat capacity of iron.

1 answer:

Anestetic [448]3 years ago

8 0

Answer:

0.39 J/g°c

Explanation:

= heat / unit of mass × unit of temperature

986.75J/16.75g

= 58.9 J/g

∆T=175°c - 25°c = 150°c

986.75 / 150°c = 6.578

986.75 / 16.75g.150°c = 0.30 j/g°c

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Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

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you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point

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Equal amounts of water are kept in a cap and in a dish which will evaporate faster ?why

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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