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Vinil7 [7]
3 years ago
9

Why does a small particle size generally cause a reaction to proceed faster?

Physics
1 answer:
Ugo [173]3 years ago
7 0
I think its a. i am not sure though.
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i'm begging ya'll i posted this about 3 times already and i'm really stuck. can someone please give me the answers to the cart?
Vedmedyk [2.9K]

Answer:

just put the molecules and atoms where they belong

Explanation:

4 0
2 years ago
A particle of mass moves under a force given bywhere and are unit vectors in the and directions. The particle is placed at the o
krek1111 [17]

The elapsed time when the particle returns to the origin is determined from the ratio of initial velocity and acceleration of the particle.

<h3>Time of motion of the particle</h3>

The time of motion of the particle is calculated by applying Newton's second law of motion.

F = ma

F = m(v)/t

where;

  • t is time of motion of the particle
  • m is mass of the particle
  • v is velocity of the particle

a = v - u/t

v = u + at

when the particle returns to the origin, direction of u, = negative.

final velocity = 0

0 = -u + at

at = u

t = u/a

Learn more about force here: brainly.com/question/12970081

#SPJ11

4 0
1 year ago
Calculate the momentum of a 6 kg rock that is rolling down a hill with a velocity of 4 m/s.
Leokris [45]

Answer:

24kgm/s

Explanation:

momentum(p)=mass×velocity

=6×4

=24 kgm/s

4 0
3 years ago
What type of image is formed by a mirror if m= -6.1?
mrs_skeptik [129]
I Think The answer is a I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you
5 0
2 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
2 years ago
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