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liberstina [14]
3 years ago
12

a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's

Physics
1 answer:
Andrew [12]3 years ago
7 0

Answer:

53 m/s^2

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

a = \frac{v^2}{r}

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

a = \frac{525^2}{5200} = 53m/s^2

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ivolga24 [154]
The origin of the universe
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2 years ago
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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
2 years ago
Which electromagnetic waves have the shortest wavelengths and highest frequencies?
Usimov [2.4K]

Answer: Gamma rays

Explanation: The given waves belong to the electromagnetic spectrum which consists of different electromagnetic radiations arranged in terms of increasing wavelengths or decreasing frequencies.

E= h\times \nu

\nu=\frac{c}{\lambda}

Thus E=\frac{h\times c}{\lambda}

E= energy

\nu= frequency

c = speed of light

\lambda = wavelength

Thus frequency and wavelength are inversely related. The waves having high energies ave high frequencies and have shorter wavelengths.

Thus gamma rays having highest energy have highest frequency and shortest wavelength.


3 0
2 years ago
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A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c
natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

mg = \rho V_{displaced} g

V_{displaced} = \frac{m}{\rho}

V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

V\rho_w = m_{ice}

V = \frac{0.200}{1000} = 2\times 10^{-4} m^3

So here extra volume that rise in the level will be given as

\Dleta V = V - V_{displaced}

\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}

(\pi (0.039^2) h = 0.41 \times 10^{-4}

h = 0.86 cm

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0
3 years ago
What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co
MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0
2 years ago
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