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PolarNik [594]
2 years ago
14

A 10.0 kg crate is pushed with a horizontal force of 40 N. The crate moves at a constant velocity of 3.0 m/s. What is the value

of the friction force on the crate?
Chemistry
1 answer:
irga5000 [103]2 years ago
5 0

Answer:

40 N

Explanation:

Newton's second law of motion states that the net force acting on the crate is equal to the product between its mass and its acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have:

- The net force can be written as

\sum F = F - F_f

where

F = 40 N is the horizontal forward push

F_f is the force of friction

Also, we know that the crate moves at constant velocity, so its acceleration is zero:

a=0

By combining all the equations together, we find:

F-F_f=0\\F_f = F = 40 N

So, the force of friction is 40 N.

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In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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Using the dilution equation:

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                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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