Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV=C, where C is a constant. Suppose that at a certain instant the volume is 600cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

Explanation:

The given data is as follows.

             height (h) = 4.70 m,    mass = 81.0 kg

              t = 1.84 s

As formula to calculate the velocity is as follows.

            = 2gh

                       =

                       = 92.12

As relation between force, time and velocity is as follows.

                     F =

Hence, putting the given values into the above formula as follows.

                  F =

                     =

                     = 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

Answer:

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

Answer:

2258.4 m

Explanation:

Distance covered is a product of speed and time hence

s=vt where s is the displacement/distance covered, v is the speed and t is the time taken

s=24*94.1=2258.4 m

Therefore, the distance covered is 2258.4 m

In an electric circuit, resistance and current are ____

A. directly proportional

B. inversely proportional

C. have no effect on each other

Explanation:

A

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

        W = mg

let's use the concept of density

        ρ_body = m / V

        m = ρ_body V

        W = ρ_body V g

The thrust of the body is given by Archimedes' law

        B = ρ_liquid g V_liquid

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

       ρ_body V g = ρ_liquid g V_liquid

       ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.