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3 years ago

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All of these are examples of chemical reactions except _____________________________.

1 answer:

Luda [366]3 years ago

4 0

C. The heating element on a stove turning from black to orange

Hope this helps =]

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The weight of a standard object defined as having a mass of exactly 2.9 kg is measured to be 28.449 n. in the same laboratory, a

PtichkaEL [24]

Mass of the object m = 2.9 kg
Force F1 = 28.449 N
F1 = m1 x a => a = F / m => 28.449 / 2.9 => a = 9.81, which is gravitational acceleration.
In the same lab, a = g = 9.81, second object F2 = 48.7N = m2 x a
m2 = F2 / a => 48.7 / 9.81 => m2 = 4.96 kg
Mass of the second object m2 = 4.96 kg

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3 years ago

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Describe how two isotopes of nitrogen differ from two nitrogen ions???PLEASE ANSWER 20 POINTS

Wewaii [24]

Nitrogen isotopes don't have a charge.

3 0

3 years ago

During a thunderstorm, a plasma channel forms between the negatively charged clouds and the positively charged Earth, which crea

olchik [2.2K]

Electrons would likely pass through the plasma channel when the formation of a lightning takes place. Lightning is primarily caused by the formation of an ionised cloud in the atmosphere. This cloud was due to the attractions between molecules of rain drops wherein produce a negative charge.

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3 years ago

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If the atmospheric pressure in a tank is 23 atmospheres at an altitude of 1,000 feet, the air temperature in the tank is 700F, a

liraira [26]

Answer:

W = 289.70 kg

Explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

$\frac{1}{n} = \frac{RT}{PV}$

$= \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}$

$= 1.02\times 10^{-4}$

$n = 10^4 mole$

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is $W = 10^4\times 28.91 g = 289700 g$

W = 289.70 kg

6 0

3 years ago

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe agai

QveST [7]

Answer:

Explanation:

Given mass of grindstone $m=90\ kg$

radius of stone $r=0.34\ m$

angular speed of disc $\omega =90\ rpm$

Steel Axle applying a force of $F=20\ N$

coefficient of kinetic friction $\mu =0.2$

Frictional Torque applied by steel is given by

$\tau=r\times f_r$

$\tau =r\times \mu F$

where $f_r$ =frictional force

$\tau =r\times \mu \times F$

$\tau =0.34\times 0.2\times 20$

$\tau =1.36\ N-m$

Torque is also given by

$\tau =I\cdot \alpha$

where $\alpha$ =angular acceleration

I=moment of Inertia

$\tau =0.5Mr^2\times \alpha$

$0.5Mr^2\times \alpha =1.36\ N-m$

$0.5\times 90\times 0.34^2\times \alpha =1.36$

$\alpha =0.261\ rad/s^2$

3 0

4 years ago