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3 years ago

What happens to volume if pressure and temperature are doubled?

Physics

1 answer:

Ahat [919]3 years ago

8 0

Pv=nrt if the temp. and pressure double the equation is sound and volume remain

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Two forces of 50 N and 30N respectively, are acting on an object. Find the net force (in N) on the object if

xxTIMURxx [149]

A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions

Answer:

A) 80 N

B) 20 N

Explanation:

A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..

Thus;

ΣF = 50 + 30

ΣF = 80 N

B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;

ΣF = 50 - 30

ΣF = 20 N

8 0

3 years ago

A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container

Natalka [10]

Given:

A cylindrical container closed of both end has a radius of 7cm and height of 6cm.

Explanation:

A.) Find the total surface area of the container.

A = 2πrh + 2πr²
A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
A = 263.76 + 307.72
A = 571.48

B.) Find the volume of the container.

V = πr²h
V = (3.14)(7×7)(6)
V = 923.16

Not sure huhuness.

#CarryOnLearning

8 0

3 years ago

Read 2 more answers

12000 inches to yards

Nutka1998 [239]

ANSWER

$\begin{equation*} 333.33\text{ yds} \end{equation*}$

EXPLANATION

We want to convert 12000 inches to yards.

To do this, divide the value in inches by 36:

$\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}$

That is the answer.

4 0

1 year ago

A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the

levacccp [35]

Tension in the rope due to applied force will be given as

$F = 142 N$

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

$W = F d cos\theta$

now we can plug in all values above

$W = 142 * 6.1 * cos37$

$W = 691.8 J$

So here work done to pull is given by 691.8 J

8 0

3 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

3 years ago