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2 years ago

14

Give at least two examples where people commonly use temperature to control reaction rates.

1 answer:

wariber [46]2 years ago

7 0

The examples may be;
making yogurt; If you heat beyond 130F you kill the cultures. The higher the temperature up until that point the faster it cultures until you close in on killer temperature.
Mixing 2 part epoxy; Keep it cool and the reaction time is delayed. On the other hand, heat it up and epoxy reacts quicker. In the winter time and close to 35 Degrees, the reaction time for cement to harden can take 400% more time to set.

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A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an

Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

$\omega = \alpha t = 2.1 t$

And so the radial acceleration is

$a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2$

The tangential acceleration is always the same since angular acceleration is constant:

$a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2$

For these 2 quantities to be the same

$a_r = a_t$

$0.441 t^2 = 0.21$

$t^2 = 0.21/0.441 = 0.4762$

$t = \sqrt{0.4762} = 0.69 s$

6 0

3 years ago

Nathalie leaves a history classroom and walks 3 meters North to drinking fountain. Then she turns and walks 10 meters south to a

pishuonlain [190]

Answer:

13 meters

Explanation:

Step one:

given

We are told that Nathalie leaves a history classroom and walks 3 meters North

Then travels another 10 meters south to an art classroom.

Required

The total distance.

Step two:

The total distance can be computed by summing up the 3 meter North distance traveled and the 10 meter south distance traveled

Total distance= 3+10= 13meters

7 0

3 years ago

Can You Please Give Me Some Examples of Density-Independent and Density-Dependent Limiting Factors?

ivolga24 [154]

Density-Dependent:
1<span><span><span><span>. </span>competition.</span><span>
<span>2. </span>overcrowding.</span><span>
3<span>. </span>predators.</span></span><span>

(These are a few from a test I took, hopefully they help you a bit >.<)</span></span>

7 0

3 years ago

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

2 years ago

I need help<br><br>please help me with this

Furkat [3]

Djdjjdjdjdjjfjfhshdhehe

3 0

2 years ago