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3 years ago

What are the variables of Boyle's law?

Physics

1 answer:

vesna_86 [32]3 years ago

3 0

Answer: pressure and volume

Explanation:

Boyle's law states that the volume of a fixed mass of gas is inversely proportional to the pressure of the gas provided temperature is kept constant.

Mathematically, Boyle's law can be explained as: PV = k,

where P is pressure

V is volume, and

k is a constant.

So, increase in the pressure of a gas leads to decrease in its volume and vice versa.

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A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0

zimovet [89]

At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s

5 0

3 years ago

Paco pulls a 67 kg crate with 738 N and of force across a frictionless floor 9.0 M how much work does he do in moving the crate

olga_2 [115]

Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

5 0

2 years ago

During which interval is the object not moving

Bezzdna [24]

Answer:

Between 2.0 s and 4.0 s (B and C)

Between 5.0 s and 8.0 s (D and E)

Between 10.0 s and 11.0 s (F and G)

Explanation:

The graph shown in the figure is a velocity-time graph, which means that:

- On the x-axis, the time is plotted

- On the y-axis, the velocity is plotted

Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:

Between 2.0 s and 4.0 s (B and C)

Between 5.0 s and 8.0 s (D and E)

Between 10.0 s and 11.0 s (F and G)

From the graph, it would be possible to infer additional information. In particular:

- The area under the graph represents the total distance covered by the object

- The slope of the graph represents the acceleration of the object

6 0

2 years ago

A soccer ball is kicked 8 m north. Then a teammate kicks it 6 m east into the goal. What is the soccer ball’s displacement from

dimulka [17.4K]

Answer:

the displacement is zero and the distance is 100 meters

Explanation:

7 0

2 years ago