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olya-2409 [2.1K]

3 years ago

14

What effect does a catalyst have on a system in equilibrium?

1 answer:

Jet001 [13]3 years ago

8 0

ANSWER:

What effect does a catalyst have on a system in equilibrium?

The system is unaffected.

~batmans wife dun dun dun....

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Why does a vibrating simple pendulum not to produce any sound

Gelneren [198K]

It does produce 'sound' ... a compression wave traveling through the air. But your ears don't hear a sound that's vibrating less than 20 or 30 times every second. If you could swing your pendulum that fast, you could hear the sound of its vibrations pushing the air around.

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3 years ago

If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has

zavuch27 [327]

Answer:

1250

Explanation:

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2 years ago

A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?

GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

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3 years ago

What are used for manufacturing paper along with wood and chemicals

Tju [1.3M]

Answer:

Paper and pulp are made from cellulosic fibers and other plant materials. Some synthetic materials may be used to impart special qualities to the finished product. Paper is made from wood fibers, but rags, flax, cotton linters, and bagasse are also used in some papers.

8 0

3 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago