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3 years ago

As pressure increases, temperature must ______________ for water to remain in a gaseous state.

Physics

1 answer:

Katyanochek1 [597]3 years ago

6 0

As the pressure goes up, the temperature also goes up, and vice-versa.
<span>Also same as before, initial and final volumes and temperatures under constant pressure can be calculated.</span>

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A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t

zzz [600]

The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
$J=\Delta p=m \Delta v$
where
m=225 kg is the mass of the boat
$\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s$ is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
$J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s$
and the negative sign means the direction of the impulse is against the direction of motion of the boat.

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3 years ago

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A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n

RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

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2 years ago

If grade 7 students learn six periods and if one period is 40 minutes then how many seconds they learn per day

Ludmilka [50]

Answer:

C. 14400

Explanation:

40min × 6 periods = 240mins total.

1 minute = 60 seconds.

240 minutes × 60 = 14400 seconds.

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3 years ago

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A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

$V=v*A\\A=\pi *r^{2}$

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

$A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\$

Therefore the flow is:

$V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]$

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

$v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]$

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3 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0

3 years ago