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3 years ago

Which option would it be

Physics

1 answer:

guapka [62]3 years ago

7 0

It would be 4 atm, because the way to figure out the final pressure is that (P1)(V1)=(P2)(V2)

meaning that the original pressure x original volume is equal to the final pressure x final volume. This gas law is called Boyle's law if you'd like to learn more about it.

But (1 atm)(40 mL)=(4 atm)(10 mL)

So it would be the second choice.

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Ariana is accelerating her car at a rate of 4.6 m/s2 for 10 seconds. Her starting velocity was 0 m/s.

Oksanka [162]

answer is 46 m/s

hope it help you

3 0

3 years ago

I meed help with these 2 questions plz

ludmilkaskok [199]

-- 30N

-- the sum of all the forces

4 0

3 years ago

What is the idea behind the law of multiple proportions?

Akimi4 [234]

Answer:

The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed

Explanation:

5 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the

gregori [183]

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

beaker A

W_total = W_ empty + W_water

Beaker B

W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

4 0

3 years ago