1. The chemical reaction produced by Carlo's fire is exergonic because energy is "going out". As the reaction proceeds, entropy increases as the energy stored in the dry wood and leaves are used up as fuel to create the fire which produces low quality light and warmth.
2. This reaction is a classic example of an exothermic reaction. Exothermic reactions are characterized with the presence of heat and light in the products. Combustion reactions are always exothermic in nature.
3. Catalyst are substances that are used to speed up reactions by lowering the activation requirement. Catalysts aren't consumed in the reaction and can still be chemically retrieved afterwards. In this situation, the leaves cannot be retrieved after the reaction ends. The leaves speed up the heating of the wood but it does not behave as a catalyst.
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
The choices are confusing. Air, oil, and alcohol are fluids at any reasonable temperature. Dry cement is not.
Answer:
I would say that is false. Science can only be perfect after at least some sort of scientific communication and interaction.
Answer:
the net toque is τ=8.03* 10⁻⁴ N*m
Explanation:
Assuming the disk has constant density ρ, the moment of inertia I of is
I = ∫r² dm
since m = ρ*V = ρπR² h , then dm= 2ρπh r dr
thus
I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2
replacing values
I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²
from Newton's second law applied to rotational motion
τ= Iα , where τ=net torque and α= angular acceleration
since the angular velocity ω is related with the angular acceleration through
ω= ωo + α*t → α =(ω-ωo)/t = (21 rad/s-0)/0.8 s = 26.25 rad/s²
therefore
τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m
