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Sladkaya [172]
4 years ago
14

Consider the chemical equation CH4 + 2 O2 → CO2 + 2 H2O. In this equation, CH4 is a

Physics
2 answers:
Reika [66]4 years ago
6 0

Answer:

\huge \boxed{\mathrm{B. \ reactant}}

\rule[225]{225}{2}

Explanation:

\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O

Balancing the Hydrogen atoms on the right side,

\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O

Balancing the Oxygen atoms on the left side,

\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O

Reactants are on the left side:

\sf \boxed{\sf CH_4 + O_2} \Rightarrow CO_2 + H_2O

\rule[225]{225}{2}

Dennis_Churaev [7]4 years ago
3 0

CH4+2O2=\ \textgreater \ CO2+2H2O
Since t<span>erms which are placed left to the arrow mark are reactants  and those placed right are products you can easily solve the task represented above. According to the information I've provided CH4 is reactant and it is </span><span>Fule. So, the correct answer is B.</span>
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Fudgin [204]
A calculator must be used. To put your calculator in degree mode, press the MODE button and select degree, the press the 2nd button then MODE again. For most TI calculators, press the 2nd button then press the cos button then enter the value 0.34. This will give you an answer of 70.123 (when you round to 3 decimal places).

The answer is C
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3 years ago
A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Mariulka [41]

Answer:

The answer is 312.5j

Explanation:

The kinetic energy (KE):

KE=1/2*m*v^2

M= mass of the object

v= velocity of the object

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3 years ago
Why does a cheetah run really fast?
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3 years ago
A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 ◦
Readme [11.4K]

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, \theta=26.1^{\circ}

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, g=9.8\ m/s^2

Let N is the normal force exerted on the cart by the floor. It is given by :

N=mg-T\ sin\theta

T\ cos\theta=\mu N

T\ cos\theta=\mu (mg-T\ sin\theta)

T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))

T = 103.23 N

So, the normal force is :

N=20.4\times 9.8-103.23\ sin(26.1)

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

4 0
3 years ago
Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all
Alexandra [31]

Answer:

\Delta h=0.0364\ m=3.64\ cm

Explanation:

Given:

  • change in stiffness constant of the spring on replacing the original springs, \delta k=5355\ N.m^{-1}
  • mass of the car, m=1455\ kg
  • initial length of the original car-spring before compression, l_i=12\ cm=0.12\ m
  • final length of the original car-spring after compression, l_f=8.55\ cm=0.0855\ m

So, weight of the car:

w=m.g

w=1455\times 9.81

w=14273.55\ N

<u>Now the spring constant of original spring:</u>

w=4k_o.(l_i-l_f) (since 4 springs are in parallel)

14273.55=4k_o\times (0.12-0.0855)

k_o=103431.522\ N.m^{-1}

<u>So the stiffness constant of the new springs:</u>

k_n=k_o-\delta k

k_n=103431.522-5355

k_n=98076.522\ N.m^{-1}

<u>Now the height lowered:</u>

w=k_n.4\Delta h (since 4 springs are in parallel)

14273.55=4\times98076.522\times \Delta h

\Delta h=0.0364\ m=3.64\ cm

5 0
4 years ago
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