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solong [7]
3 years ago
15

A substance that accelerates any chemical reaction but is not consumed in the reaction is a _____.

Physics
2 answers:
Mandarinka [93]3 years ago
8 0
A. Catalyst. ---------------
dalvyx [7]3 years ago
7 0

A. Catalyst. ---------

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How did scientists discover the Earth had a liquid outer core and solid inner core?
viktelen [127]
Dr. Inge discovered the make up of the earths inner core by studying how an earthquakes waves bounced off the core. And Inge Lehmann was studying the waves of a 1929 earthquake when she found them acting inconsistently with solid mantle crust 
hope it helps you


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4 years ago
In a compression wave, particles in the medium move
MissTica
<span>The particles through which compressional waves travel move in the same direction as the wave. This may be observed by fixing one end of a large spring and then compressing and extending the other end. The wave travels from one end to the other and the spring's parts move in the same direction.</span>
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3 years ago
Read 2 more answers
As you take your leisurely tour through the solar system, you come across a 1.45 kg rock that was ejected from a collision of as
nikklg [1K]

Answer:

1.3310\times 10^8 m/s is the rock's speed.

Explanation:

Momentum is defined as motion possessed by the moving body's mass. Mathematically it a product of mas and velocity of the body.

Momentum(P)=Mass(m)\times velocity(v)

Given : Mass of rock = m = 1.45 kg

Velocity of the rock = v

Momentum of the rock = P =1.93\times 10^8 kg m/s

1.93\times 10^8 kg m/s=1.45 kg\times v

v=\frac{1.93\times 10^8 kg m/s}{1.45 kg}=1.33\times 10^8 m/s

1.3310\times 10^8 m/s is the rock's speed.

6 0
3 years ago
Argument changes meaning because of an ambiguous word or phrase.
Virty [35]
I think the answer is D
7 0
3 years ago
A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t
Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0
3 years ago
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