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yan [13]
4 years ago
10

The weight of an object four earth radii from the center of the earth is:

Physics
1 answer:
Komok [63]4 years ago
8 0

Answer:

D. 1/16 It's weight on the surface.

Explanation:

As we know that the gravity due to earth at its surface is given as

g = \frac{GM}{R^2}

now at the distance of 4 radius of earth the gravity is given as

g' = \frac{GM}{(4R)^2}

so we can say that

g' = \frac{g}{16}

so here gravity becomes 1/16 times of the gravity due to earth on its surface

now we know that weight on the surface of earth is given as

W = mg

also the weight now at the given height is given by

W' = mg/16

so here correct answer will be

D. 1/16 It's weight on the surface.

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By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. If the mass of th
yaroslaw [1]

Answer:

The astronaut's acceleration is 155.1 times the vehicle's acceleration

Explanation:

These effects due to Newton's third law of action and reaction. Since the forces are equal but in the opposite direction and each acting on a different body. We distance that the Force is F let's calculate the acceleration of the vehicle and the astronaut  

Astronaut

        F = m_{ast}  a₁

Vehicle

        F = m_{veh} a₁

F = 555.1 m_{ast} a₂

Let's match the equation

m_{ast} a₁ = 155.1 m_{ast} a₂

a₁ = 155.1 a₂

a₁ / a₂ = 155.1

The astronaut's acceleration is 155.1 times the vehicle's acceleration

We see that even when the acceleration of the vehicle is small, there is a very high multiplicative factor.

One method to improve this situation is that the vehicles fear some small retro-rocket vehicles to reduce their acceleration. This would have a very favorable impact on the astronaut's mission.

Another method would be for the astronaut himself to have the retro-rocket and control his acceleration.

4 0
4 years ago
A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, w
Travka [436]

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

emf=-N\frac{\Delta \Phi}{\Delta t}         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

emf=-(60)(\frac{-0.109W}{0.230s})=28.43V

The induced emf is 28.43V

(c) The induced current in the coil is given by:

I_{in}=\frac{emf}{R}      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

7 0
4 years ago
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence
Lynna [10]

Answer:

1) Recollapsing universe

2) critical universe

3) Coasting universe

Explanation:

According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are

1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

3) Coasting universe -  in this, expansion of universe is steady and uniform

6 0
3 years ago
A photon has momentum of magnitude 8.30×10−28 kg⋅m/s . Part APart complete What is the energy of this photon? Give your answer i
d1i1m1o1n [39]

Answer with Explanation:

We are given that

Momentum of photon=8.3\times 10^{-28} kg.m/s

a. We have to find the energy of this photon.

Speed of photon=c=3\times 10^8 m/s

We know that

Momentum=p=\frac{h}{\lambda}

Where

h=6.63\times 10^{-34}J-s=Plank's constant

\lambda=Wavelength of photon

\lambda=\frac{h}{p}

\lambda=\frac{6.63\times 10^{-34}}{8.3\times 10^{-28}}

\lambda=7.99\times 10^{-7} m

E=\frac{hc}{\lambda}

E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{7.99\times 10^{-7}}

E=2.49\times 10^{-19} J

Hence, the energy of photon=2.49\times 10^{-19} J

B.Energy of photon in electron volt=\frac{2.49\times 10^{-19}}{1.6\times 10^{-19}}=1.55 eV

Energy of photon=1.55eV

C.Wavelength of photon =\lambda=7.99\times 10^{-7}m

6 0
4 years ago
NBAのバスケットボールコートのサイズは
Marysya12 [62]

28.65m x 15.24m

Explanation:

The size of a basketball court in the NBA is usually 28.65m x 15.24m giving a total court area of 436.62m².

The NBA is the acronym for National Basketball Association. It comprises of several teams from across United states and Canada in basketball league.

  • The game is played by hand with a target basket.
  • For each successive throw into the basket, 1 or 2 or 3 points can be awarded.
  • In feets the dimensions are 94 by 50 feet
  • From the dimension we can accurately say that a basketball court is rectangular in shape.
  • James Naismith is credited with the invention of the game of basketball

learn more:

Basketball jump brainly.com/question/11222255

#learnwithBrainly

6 0
3 years ago
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