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netineya
12 days ago
12

You are designing a spacecraft intended to monitor a human expedition to Mars (mass 6.42×1023kg, radius 3.39×106m). This spacecr

aft will orbit around the Martian equator with an orbital period of 24.66 h, the same as the rotation period of Mars, so that it will always be above the same point on the equator. What must be the radius of the orbit?
Physics
1 answer:
Softa [2.9K]12 days ago
3 0
The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.
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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [3340]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
1 month ago
A 15.0 cm object is 12.0 cm from a concave mirror that has a focal length of 4.8 cm. Its image is 8.0 cm in front of the mirror.
kicyunya [3154]
The correct answer is -10.
7 0
13 days ago
Read 2 more answers
A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
Keith_Richards [3141]

Response:

Q = 8,345 * v

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Q = 8,345 * v

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

8 0
1 month ago
A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a
Maru [3252]
The response is 176 minutes. The translation of 456 g equals 0.456 kg. The specific heat of ice is 2093 J kg⁻¹, used to calculate heat required for a 25-degree rise, determined by mass multiplied by specific heat and temperature increase. The necessary calculations yield a total heat load of 176164 J. Finally, by dividing heat required by heat supply rate, we ascertain that it will take approximately 176.16 minutes.
4 0
8 days ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
Softa [2930]
B) 14.0 N

To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Now, utilizing the known data, we compute the energy prior and post.

Before:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (2.25 m/s)^2

E = 6.75 kg * 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (1.2 m/s)^2

E = 6.75 kg * 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”
8 0
17 days ago
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