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liraira [26]
3 years ago
8

Carlos was camping and getting cold as the sun went down. He wanted to light a fire for warmth and light. However, he discovered

that the nearby wood was wet, and would have no light. He had to look under some leaves and debris to find dry wood. He piled the wood and surrounded it with a circle of stones to keep the fire contained. Then he put some dry leaves around the logs to help the fire to get started. He lit the leaves. Soon the leaves had burned away, his fire was burning nicely, and Carlos was getting warmer.
1. Is the chemical reaction produced by Carlos's fire exergonic or endergonic?

2. Is the reaction in Carlos's fire endothermic or exothermic? Explain

3. Do you think the leaves were a catalyst for the fire? Why or why not?
Physics
1 answer:
storchak [24]3 years ago
6 0
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. 
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vodomira [7]
5207 because I said so
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Two spheres having masses M and 2M and radii R and 3R, respectively, are released from rest when the distance between their cent
Andrei [34K]

Answer:

v_2 = \sqrt{\frac{GM}{3R}}

v_1 = 2\sqrt{\frac{GM}{3R}}

Explanation:

As we know by energy conservation that change in gravitational potential energy of the system = change in kinetic energy of the two ball

So here we can say

-\frac{GM(2M)}{12R} + 0 = -\frac{GM(2M)}{4R} + \frac{1}{2}Mv_1^2 + \frac{1}{2}(2M)v_2^2

Also since there is no external force on the system of two masses so here total momentum of the two balls will remains conserved

0 = Mv_1 + 2Mv_2

v_1 = -2v_2

now we have

\frac{GM^2}{2R} - \frac{GM^2}{6R} = \frac{1}{2}M(-2v_2)^2 + \frac{1}{2}(2M)v_2^2

\frac{GM^2}{3R} = Mv_2^2

v_2 = \sqrt{\frac{GM}{3R}}

v_1 = 2\sqrt{\frac{GM}{3R}}

4 0
4 years ago
FREE BRAINLIST need help solving this​
Mademuasel [1]

Answer: 400

Explanation

4 0
3 years ago
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Explain why ice melts in a cup at room temperature, in terms of energy transfer and the second law of thermodynamics.
zlopas [31]

Answer:

The second law of thermodynamics describes the direction in which heat is transferred between systems, <u>heat is a form of energy</u> in transition.

This law says that heat or energy always flows spontaneously from the body or system with a higher temperature to a lower temperature system (from something hot to something cold, and not the other way around).

This is why if we leave an ice at room temperature it will eventually melt, because <u>the environment transferred energy to the ice</u> and caused its temperature to increase and thus to turn into a liquid form.

The second law can also be interpreted in terms of entropy, and tells us that entropy, which is often interpreted as a measure of disorder, always increases.

4 0
4 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

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