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2 years ago

At what stage of a reaction do atoms have the highest energy?

1 answer:

3 0

The correct answer for the question that is being presented above is this one: "c. transition state stage." During the transition state stage, the reaction of the atoms have the highest energy. It is also <span>during the formation of the activated complex in the middle of the experiment.</span>

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What is 178.57 rounded to the nearest whole number

postnew [5]

<h2><em>THANK</em><em> </em><em>YOU</em><em> </em></h2>

4 0

2 years ago

Read 2 more answers

supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan

saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

7 0

3 years ago

Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t

padilas [110]

Answer:

$\omega=32.14\ rad/s$

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

$v=\omega\times A$

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

$\omega=\dfrac{v}{A}$

$\omega=\dfrac{2.08\ m/s}{0.0647\ m}$

$\omega=32.14\ rad/s$

So, the angular frequency of the spring is 32.14 rad/s.

4 0

3 years ago

Describe three ways that adults continue to guide children's emotions during the school-age

wolverine [178]

1. Always be the bigger person
2. Violence is never the answer
3. Don’t fight fire with fire

6 0

3 years ago

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec

juin [17]

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = $= \omega^2 \times m$

k = (11.3)² m

k = 127.7 m

where;

$x_1$ = 0.065 m

$x_2$ = 0.048 m

According to the conservation of energies;

$E_1=E_2$

∴

$\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2$

$kx_1^2 = mv_2^2 + kx_2^2$

$(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2$

$0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325 - 0.2942208 = v_2^2 \\ \\ v_2^2 = 0.2453117 \\ \\ v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}$

4 0

2 years ago