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3 years ago

How can the rate of a reaction be increased?

Physics

2 answers:

VladimirAG [237]3 years ago

7 0

Answer:

A - Increasing the temperature.

Explanation:

Raising the reaction temperature by 10 °C can double or triple the reaction rate. This is due to an increase in the number of particles that have the minimum energy required. The reaction rate decreases with a decrease in temperature.

artcher [175]3 years ago

3 0

I think the answer is A- increasing the temperature.

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For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

$z = \frac{V_o}{i_o}$

now we will have

$z = \frac{125}{3.20} = 39.06 ohm$

Part b)

Now we also know that

$\frac{R}{z} = cos\theta$

$\theta = 0.982 rad = 56.3 degree$

now we have

$\frac{R}{39.06} = cos56.3$

$R = 21.7 ohm$

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This interaction is known as constructive interference. It's a result of linear superposition.

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You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

Fynjy0 [20]

Answer:

Explanation:

The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

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3 years ago

A two-stage rocket is traveling at 1200 mis with respect to the earth when the first stage runs out of fuel. Explosive bolts rel

Aloiza [94]

Answer:

The speed of the second stage after separation is 4905 m/s

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the system first stage - second stage must remain constant before and after the separation. The momentum of the system is calculated by adding the momentums of each object:

initial momentum = final momentum

m₁₊₂ · v = m1 · v1 + m2 · v2

Where:

m₁₊₂ = mass of the two stage rocket

v = velocity of the rocket

m1 = mass of stage 1

v1 = velocity of stage 1

m2 = mass of stage 2

v2 = velocity of stage 2

We have the following data:

m1 = 3 · m2

m₁₊₂ = m1 + m2 = 3 · m2 + m2 = 4 · m2

v = 1200 m/s

v1 = -35 m/s (let´s consider the backward direction as negative)

v2 = ?

Then, replacing these data in the equation of momentum of the system:

m₁₊₂ · v = m1 · v1 + m2 · v2

4 m2 · 1200 m/s = 3 m2 · (-35 m/s) + m2 · v2

Let´s solve the equation for v2:

divide both sides of the equation by m2:

4 · 1200 m/s = 3 · (-35 m/s) + v2

4800 m/s = -105 m/s + v2

v2 = 4800 m/s + 105 m/s = 4905 m/s

The speed of the second stage after separation is 4905 m/s

6 0

3 years ago

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