answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rufina
2 months ago
13

A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

at would keep the cable taut. A motorized climber could slowly carry rockets to the top, where they could be launched away from the earth using much less energy.
What would be the escape speed for a craft launched from a space elevator at a height of 56,000 km?
Physics
1 answer:
Softa [3K]2 months ago
0 0
The escape speed required for the craft is 1.49 m/s. In this scenario, we are calculating the escape velocity for a vehicle launched from a space elevator situated at a height of 56,000 km. The escape velocity can be determined using the formula provided, where G represents the universal gravitational constant and M corresponds to the mass of the Earth. The distance d can be expressed as r + h, where r is the Earth's radius. Thus, the determined escape speed is 1.49 m/s.
You might be interested in
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3153]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
2 months ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
Ostrovityanka [3204]
The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

     The distance the electron covers is  x = 2.50 \ m

Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

      F = 1.52 *10^{-16} \ N

Generally, the work-energy theorem is mathematically framed as

          W = \Delta KE

Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

        W = F* x *cos( \theta )

Where  \theta = 0 ^o assuming that the electron's movement aligns with the x-axis  

        So

             \Delta KE = F * x cos (0)

Inserting values

         \Delta KE = 1.52 *10^{-16} * 2.50 cos (0)

          \Delta KE = 3.8*10^{-16} J

According to the conservation of energy

       \Delta PE = - \Delta KE

Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

7 0
2 months ago
The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0
2 months ago
Read 2 more answers
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
serg [3582]

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

6 0
3 months ago
Read 2 more answers
Other questions:
  • A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
    5·2 answers
  • In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
    10·2 answers
  • Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti
    6·1 answer
  • Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus
    10·1 answer
  • briefly discuss the phenomenon of childhood maltreatment in the context of five different psychological perspectives
    8·1 answer
  • Blue light, which has a wavelength of about 475 nm, is made to pass through a slit of a diffraction grating that has 425 lines p
    12·2 answers
  • For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
    13·1 answer
  • In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6
    7·1 answer
  • Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b
    5·2 answers
  • A 5.00 kilogram mass is traveling at 100 meters per second. Determine the speed of the mass after an impulse of 30 Newton * seco
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!