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3 years ago

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In a standard gas grill propane tank, there is approximately 4,579 mL of propane (C3H8). At a temperature of 55˚C, the tank has

a pressure of 1,798 kPa. The tank is cooled to 25˚C and the pressure reduced to 1,025 kPa. What will the new volume be?

1 answer:

AnnyKZ [126]3 years ago

8 0

Answer:

7300 mL

Explanation:

The question is a bit misleading, since tanks generally have a constant volume. But anyway, using ideal gas law:

PV = nRT

If amount of gas (n) is constant, then:

PV / T = constant

P₁V₁ / T₁ = P₂V₂ / T₂

(1798 kPa) (4579 mL) / (55 + 273) K = (1025 kPa) V / (25 + 273) K

V = 7298 mL

Rounded to two significant figures, the volume is 7300 mL.

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A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?

Anuta_ua [19.1K]

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

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The apparent backward motion of a planet along the celestial sphere is called ____________ motion.

motikmotik

Answer:

Retrograde

Explanation:

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3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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4 years ago

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tatyana61 [14]

Answer:

counting tree rings can indicate age

Explanation:

hope i helped at least a little bit :)

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Answer:

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