A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 600lb, what is the size of the force when it is at each of the following distances from the Earths surface?a. 20,000 miles
b. 30,000 miles
c. 100,000 miles

Answers

Answer 1
Answer: Grav Force = GMm/r squared. M is earth's mass. m is the probe. r is distance to centre of earth.Given that 600 = GMm/(10,000^2).GMm = 600x(10,000^2) for 10, 000 milesGMm = ? x (20,000^2) for 20, 000 milestherefore600x(10,000^2)=? x (20,000^2)therefore [600x(10,000^2)]/(20,000^2)=? [600x(1x1)]/(2x2)=? [600/4]=? ?=150lb.Same procedure for b and c.Please ask if you want further help.
Answer 2
Answer:

gravitational force varies based on 1/r^2

when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.

As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)




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How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature? A. 334,584 J B. 167,292 J C. 100,375 J D. 450,759 J

Answers

Answer:

Option C is the correct answer.

Explanation:

Heat required to melt solid in to liquid is calculated using the formula

            H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

          H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Express your answer in micrometers(not in nanometers).

Answers

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=(y)/(D)\n\Rightarrow \theta=tan^(-1){(y)/(D)}\n\Rightarrow \theta=tan^(-1){(4.84* 10^(-3))/(3)}\n\Rightarrow \theta=0.09243\ ^(\circ)

sin\theta=(\lambda)/(d)\n\Rightarrow d=(\lambda)/(sin\theta)\n\Rightarrow d=(598* 10^(-9))/(sin0.09243)\n\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=(\lambda')/(2)\n\Rightarrow \lambda'=2dsin\theta\n\Rightarrow \lambda'=2* 0.00037066* sin0.09243\n\Rightarrow \lambda'=1.196* 10^(-6)\n\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

Final answer:

The wavelength of light that will produce the first-order dark fringe at the same point on the screen is the same as the original wavelength of the light, which is 598 nm (0.598 μm).

Explanation:

To find the wavelength of light that will produce the first-order dark fringe at the same point on the screen, we can use the equation dsinθ = nλ, where d is the separation between the slits, θ is the angle of the fringe, n is the order of the fringe, and λ is the wavelength of the light.

In this case, the first-order bright fringe is located at a distance of 4.84 mm from the center of the central bright fringe. Since this is a first-order fringe, n = 1.

Plugging in the values, we have (0.120 mm)(sinθ) = (1)(λ). Rearranging the equation gives sinθ = λ/0.120 mm.

Since the first-order dark fringe is located at the same point as the first-order bright fringe, the angle of the first-order dark fringe can be calculated by taking the sine inverse of λ/0.120 mm.

Finally, to find the wavelength of light that will produce the first-order dark fringe at this point, we can rearrange the equation to solve for λ: λ = (0.120 mm)(sinθ).

Now, substitute the known values into the equation to calculate the wavelength of light:

λ = (0.120 mm)(sinθ) = (0.120 mm)(sin sin^-1(λ/0.120 mm)) = λ.

The wavelength of light that will produce the first-order dark fringe at this point on the screen is the same as the original wavelength of light, which is 598 nm. Converting this value to micrometers, we get 0.598 μm.

Learn more about interference pattern here:

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A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed v = 2.20*10^7\ m/s

Acceleration a=1.90*10^(13)\ m/s^2

We need to calculate the magnetic field

Using formula of magnetic field

F=qvB....(I)

Using newton's second law

F= ma....(II)

From equation (I) and (II)

ma=qvB

Put the value into the formula

1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B

3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B

B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))

B=0.009014\ T

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

A sound wave travels with a velocity of 330 m/s and has a frequency of 500 Hz. What is itswavelength?

Answers

Answer:

Wavelength = 0.66 meters

Explanation:

Given the following data;

Speed = 330 m/s

Frequency = 500 Hz

To find the wavelength;

Mathematically, wavelength is calculated using this formula;

Wavelength = \frac {speed}{frequency}

Substituting into the equation, we have;

Wavelength = \frac {330}{500}

Wavelength = 0.66 meters

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?A. F

B. 4F

C. 4F/3

D. 4F/9

E. F/3

Answers

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=(kQ^2)/(r^2) ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=(kQ* 4Q)/((3r)^2)\n\nF'=(4kQ^2)/(9r^2).....(2)

Dividing equation (1) and (2), we get :

(F)/(F')=((kQ^2)/(r^2))/((4kQ^2)/(9r^2))\n\n(F)/(F')=(kQ^2)/(r^2)* (9r^2)/(4kQ^2)\n\n(F)/(F')=(9)/(4)\n\nF'=(4F)/(9)

Hence, the correct option is (d) i.e. " 4F/9"

Final answer:

The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.

Explanation:

The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.

Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

Answers

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.