4. A helicopter is flying at a height of 1.5km. If it has 50MJ of G.P.E., what is the mass of the

Pascal's Principle states that (a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished

nika2105 [10]

Answer:

a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished at every point in the fluid and on the walls of the container.

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

P=F/A

P1=P2

F1/ A1= F2/ A2

F2= F1* A2/ A1

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

5 0

2 years ago

True or false : Scientific theories have withstood the test of time and are accepted as proven fact.

Ilya [14]

False:Laws are theories that have not been proven false.

8 0

3 years ago

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A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.

Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0

3 years ago

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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

2 years ago