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Vadim26 [7]
3 years ago
9

How does temperature affect chemical reaction​

Physics
1 answer:
Elis [28]3 years ago
8 0
When its temperature increases, the reactants have more kinetic energy so the frequency of effective collision increases, resulting in a faster rate of chemical reaction.
You might be interested in
Calculate the time period of simple pendulum whose length is 98.2cm
alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \  \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

5 0
1 year ago
Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di
vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

A = \pi (\frac{D}{2})^2

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

d = \pi r^2 e

r^2_e= \frac{d}{\pi}

r_e = \sqrt{\frac{d}{\pi} }

replacing d = \frac{\pi (\frac{D}{2})^2 }{10, 000} in the equation above; we have:

r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }

r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}

r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}

r_e = 500 ly

The distance (s) between each civilization = 2(r_e)

= 2 (500 ly)

= 1000 light-years (ly)

4 0
3 years ago
Kuiper Belt objects are composed of which substances? Select all that apply.
levacccp [35]
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron

8 0
3 years ago
When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
3 years ago
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

<u>D = - 0.5i - 0.25j</u>

Therefore,

<u>Dx = -0.5</u>

<u>Dy = -0.25</u>

8 0
3 years ago
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