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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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3 years ago
Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st
DedPeter [7]

Answer:

E1  = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q)  = +11nC

distance between thin glass rods   = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = \frac{Q}{2\pi e_{0}rL }

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = \frac{Q}{2\pi e_{0}rL } ( \frac{1}{0.01} - \frac{1}{0.03} )    ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

   = \frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )

   = 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = \frac{Q}{2\pi e_{0}rL }( \frac{1}{0.02} - \frac{1}{0.02} )

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0
2 years ago
Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A
KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns          Nb = 610 turns

Current = 2.80 A

Average flux through  A  = 300 μWb

Average of flux through B = 90.0  μ Wb

a) L = \dfrac{N \phi}{I}

   L = \dfrac{610\times 90 \times 10^{-6}}{2.80}

   L =19.6 mH

b) inductance of A

   L = \dfrac{N_A \phi_A}{I_A}

   L = \dfrac{430\times 300 \times 10^{-6}}{2.80}

   L =46 mH

c) magnitude of the emf

    \epsilon_B = -L_B\dfrac{dI}{dT}

    \epsilon_B = -(19.6\times 10^{-3})(0.5)

    \epsilon_B = -9.8\times 10^{-3}\ V

    \epsilon_B = -9.8\ mV

7 0
3 years ago
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