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I am Lyosha [343]
3 years ago
5

The diagram below shows a person using a device called a jetpack. Water is forced downwards from the jetpack and produces an upw

ard force on the person. State the condition necessary for the person to be able to remain stationary in mid-air.

Physics
1 answer:
klasskru [66]3 years ago
5 0

Answer:

The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST

Explanation:

Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.

Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.

Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack

The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:

a) Gravitational force

b) Mass of the jet pack user

c) Mass of the jet pack itself

d) The water which serves as fuel for the jet pack.

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A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as  1.49 x 10⁸ kilometres .)

6 0
3 years ago
The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k
Brrunno [24]

Answer:

The gravitational force changing velocity is

\frac{dF}{dt}=-8\frac{N}{s}

Explanation:

The expression for the gravitational force is

F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}

Differentiate the above equation

\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}

The velocity is the distance in at time so

V=\frac{dr}{dt}=0.4 \frac{km}{s}

\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

\frac{dF}{dt}=-8\frac{N}{s}

8 0
3 years ago
A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t
Talja [164]

Answer:

P = 133.13 Watt

Explanation:

Initial angular speed of the ferris wheel is given as

\omega_i = 2\pi f

\omega_i = 2\pi(8.5/3600)

\omega_i = 0.015 rad/s

final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

\alpha = 1.27 \times 10^{-4} rad/s^2

now torque due to friction on the wheel is given as

\tau = I \alpha

\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})

\tau = 8875.3 N m

Now the power required to rotate it with initial given speed is

P = \tau \omega

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8 0
2 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
BartSMP [9]

Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

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