To calculate for the pressure of the system, we need an equation that would relate the
number of moles (n), pressure (P), and temperature (T) with volume (V). There are a number of equations that would relate these values however most are very complex equations. For
simplification, we assume the gas is an ideal gas. So, we use PV = nRT.<span>
PV = nRT where R is the universal gas
constant
P = nRT / V</span>
<span>P = 3.40 mol ( 0.08205 L-atm / mol-K ) (251 + 273.15 K) / 1.75 L </span>
<span>P = 83.56 atm</span>
<span>
</span>
<span>Therefore, the pressure of the gas at the given conditions of volume and temperature would be 83.56.</span>
<span>when an object is submerged in the water it losses its weight as accordinglyto</span>
Answer:
the normal force on the rock acts perpendicular to the bowl's surface.
Explanation:
As we know that Normal force is the reaction force of two contact surfaces which always act perpendicular to the contact surfaces
Here we know that the rock is moving inside the bowl
So Normal force on the rock must perpendicular to the surface of the bowl which always passes through the center of the bowl.
Since the rock is moving in vertical plane so it must have two acceleration
1) Tangential acceleration which will increase the magnitude of the speed along the tangential path
2) Centripetal acceleration which will change the direction of the rock
So here only correct option will be
the normal force on the rock acts perpendicular to the bowl's surface.
Answer:
4.43 kW
Explanation:
Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.
P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀
= 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A
= 4428.1 W
= 4.4281 kW ≅ 4.43 kW
Answer:
Classification
– Drug (highly regulated)
§ A substance which changes body structure or function.
§ Stimulate hormone secretions
§ Looks like medicine and/or is administered differently than foods
– Dietary Supplement (not highly regulated)
§ Highly refined products
§ No positive nutritional value
Explanation:
https://web.cortland.edu/buckenmeyerp/Lecture14.html
