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4vir4ik [10]
2 years ago
6

Select all the correct answers. which changes will increase the rate of reaction during combustion? decreasing the area of conta

ct between the reactants adding more oxygen to the reaction removing heat from the reaction changing the reactants from solid form to powdered form lowering the exposure of the reactants to air
Physics
1 answer:
Aleks [24]2 years ago
5 0

Adding more oxygen to the reaction and changing the reactants from solid form to powdered form will increase the rate of reaction during combustion.

<h3>What are factors on which rate of chemical reactions depend?</h3>

The rate of a chemical reaction is affected by a number of factors, including temperature, the concentration of the reactants, the presence of a catalyst, and others.

  • The rate of a chemical reaction rises with temperature because more reactant and product particles will collide as a result of the higher temperature.
  • By converting the reactants from a solid into a powder, more collisions between the reactant particles will occur, which will speed up the production of the products. Increasing the surface area of the reactants also increases the rate of reaction.
  • The rate of reaction will grow when the concentration of the reactants, such as increasing the amount of oxygen in the reaction that is one of the reactants, increases.

Learn more about rate of chemical reaction here:

brainly.com/question/8497438

#SPJ4

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3 years ago
Two resistors have resistances R(smaller) and R(larger), where R(smaller) &lt; R(larger). When the resistors are connected in se
just olya [345]

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

5 0
3 years ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
3 years ago
A muscle inserts 1.5 cm from the joint axis and exerts 300 N of force at an angle of pull of 60 degrees. How much torque is prod
Amanda [17]

Answer:

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance    ...................1

put here all these value in equation 1

we get torque

torque  = force × sinθ × distance

torque  = 300 × sin60 × 1.5 ×10^{-2}

torque  = 300 × 0.8660 × 1.5 ×10^{-2}

torque  = 259.80 × 1.5 ×10^{-2}

torque  = 389.711 ×10^{-2}

torque = 3.897 N-m

8 0
3 years ago
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