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adoni [48]
4 years ago
9

In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

arth at a distance 2.50×108m from the center of the earth. At what speed must this spacecraft be launched from the earth's surface? Neglect air resistance and the gravitational pull of the moon.
Physics
1 answer:
Mariulka [41]4 years ago
6 0

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

    F = m a

    G m M / x² = m dv / dt = m dv/dx  dx/dt

    G M / x² = dv/dx   v

    GM dx / x² = v dv

We integrate

    v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m  and the velocity v = vo and the upper limit x = 2.50 10⁸m  with a velocity of v = 8.50 10³ m/s

    ½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

    72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

    72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3  10⁻⁸)

    72.25 10⁶ - v₀² = -1.213  10⁸

    v₀² = 72.25 10⁶ + 1,213 10⁸

    v₀² = 193.6 10⁶

    v₀ = 13.9 10³ m / s

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You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
3 years ago
A shoe has a length of 11 inches. Which unit conversion fraction should you use to find the length in centimeters?
valkas [14]
1 inch = 2.54cm. 11*2.54 = 27.94cm. You would use multiplication. Hope this helps!! Can I get brainliest pls
6 0
3 years ago
Read 2 more answers
What are the magnifications of the objective lenses
Anvisha [2.4K]

Answer:

Objective Lenses: Usually you find 3 or 4 objective lenses on a microscope. They almost always consist of 4x, 10x, 40x and 100x powers, when coupled with a 10x (most common) eyepiece lens, total magnification is 40x (4x times 10x) 100x, 400x and 1000x

Explanation:

Objective lenses come in various magnification powers, with the most common being 4x, 10x, 40x, and 100x, also known as scanning, low power, high power and typically oil immersion objectives, respectively

7 0
3 years ago
A particle is moved along the x-axis by a force that measures 10/(1+x)^2 pounds at a point x feet from the origin. Find the work
jarptica [38.1K]

Answer:

9 ft*lb

Explanation:

super simple but you just have to understand that the integral is going with the curve

work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb

6 0
3 years ago
What is the net force needed to accelerate a 5 kg object at 3 m/s2? Suppose that in this situation you discovered that there is
Rudik [331]

The Force need to accelerate the object is by 3 m/s² is 15 N.  Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

<h3>Force:</h3>

This can be defined as the product of the mass and the acceleration of a body. The S.I unit of force is kgm/s or Newton(N)

To calculate the force needed to accelerate a mass of 5 kg object at 3 m/s² we use the formula below.

Formula:

  • F = ma........ equation 1

Where:

  • F = Net force needed to accelerate the object
  • m = mass of the object
  • a = acceleration of the object

From the question,

Given:

  • m = 5 kg
  • a = 3 m/s²

Substitute these values into equation 1

  • F = 5(3)
  • F = 15 N

Suppose a frictional force of 5 N acts on the motion, The force applied is

  • F = F'+ma............ Equation 2

Where:

  • F = Frictional force = 5N

Substitute into equation 2

  • F = 5(3)+5
  • F' = 15+5
  • F = 20 N.

Hence, The Force need to accelerate the object is by 3 m/s² is 15 N. Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

Learn more about force here: brainly.com/question/12970081

4 0
3 years ago
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