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2 years ago

If the volume of a container increases what effect will this have on pressure? Why?

Physics

2 answers:

GaryK [48]2 years ago

7 0

Answer:

In a container of the same volume, a greater number of molecules results in greater pressure. Gas pressure increases as container volume decreases.

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frozen [14]2 years ago

5 0

As the volume of the container increases the pressure inside will decrease because the atoms have more room to move around in.

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There are screws all around you. Name five examples of screws that you see in everyday life? Think broadly!

Ira Lisetskai [31]

Answer:

Some examples of the uses of a screw are in a jar lid, a drill, a bolt, a light bulb, faucets, bottle caps and ball point pens.

Explanation:

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2 years ago

What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?

Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

$p = mv$

Let's replace what we know and solve for whatever's left

$100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s$

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2 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)

From equation 2 and 3, we get

$M_{a}$ *a = $M_{b}$ $\sqrt{a^{2} +g^{2} }$

Squaring both sides we get

$M_{a} ^{2}$ * $a^{2}$ = $M_{b} ^{2}$ * ( $a^{2}$ + $g^{2}$ )

$M_{a} ^{2}$ * $a^{2}$ = ( $M_{b} ^{2}$ * $a^{2}$ )+ ( $M_{b} ^{2}$ * $g^{2}$ )

( $M_{a} ^{2}$ - $M_{b} ^{2}$ ) * $a^{2}$ = $M_{b} ^{2}$ * $g^{2}$

$a^{2}$ = $M_{b} ^{2}$ * $g^{2}$ /( $M_{a} ^{2}$ - $M_{b} ^{2}$ )

Taking square root on both sides, we get acceleration a

a = $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ )

Hence substituting the value of a in equation 1, we get

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

3 0

3 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$