A space shuttle orbits the earth at an altitude 100 km above the earth's surface, with constant speed v = 19000 km/h. Find the magnitude of the shuttle's acceleration (in km/h²) , assuming that the radius of the earth is 6378 km. (Use decimal notation. Round your answer to an integer.) a = km/h?

A space shuttle orbits the earth at an altitude 100 km above the earth's surface, with constant speed v = 19000 km/h. Find the magnitude of the shuttle's acceleration (in km/h²) , assuming that the radius of the earth is 6378 km. (Use decimal notation. Round your answer to an integer.) a = km/h?

Name: section 13.5 question 18 A space shuttle orbits the earth at an altitu
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: section 13.5 question 18 A space shuttle orbits the earth at an altitude 100 km above the earth's surface, with constant speed v = 19000 km/h. Findthe magnitude of the shuttle's acceleration (in km/h²) , assuming that the radius of the earth is 6378

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

Section: Chapter Questions

Problem 2.10CYU: Check Your Understanding Verify that vector v V obtained in Example 2.14 is indeed a unit vector by...

See similar textbooks

Similar questions

The planet Jupiter has a radius of 43000 miles and rotates one revolution every 10 hours. What is the linear speed of a point on its equator, in miles per hour? The linear speed is approximately miles per hour. (Round to the nearest integer as needed.)
A high-tech drone flies over Miami, FL in a direction 58.3° west of north with a speed of 30.2 km/h. Three hours later, the drone suddenly shifts due north, and its speed slows to 12.7 km/h. How far (in kilometers) from Miami, FL is the drone 4.50 h after it passes over the Miami. Round your answer to 0 decimal places.
The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius R of the Earth, g, and h. 8h = Suppose a 70.00 kg hiker has ascended to a height of 1815 m above sea level in the process of climbing Mt. Washington. By what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use g = 9.807 m/s² and R = 6.371 × 106 m. Enter your answer as a positive value. weight change = %
You are on a treasure hunt and your map says "Walk due west for 23.1 paces, then walk 40.0° north of west for 48.7 paces, and finally walk due north for 30.7 paces." What is the magnitude of the component of your displacement in the direction (a) due north and (b) due west? (a) Number i (b) Number i T Units Units B OB W A N S E Start
The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius R of the Earth, g, and h. &h= Suppose a 74.35 kg hiker has ascended to a height of 1893 m above sea level in the process of climbing Mt. Washington. By what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use g = 9.807 m/s² and R = 6.371 × 106 m. Enter your answer as a positive value. weight change = %
L74. A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point (Fig. 1.39). Find the magnitude Figure 1.39 Problem 1.74. STAKT FINISH 5.80 km- Jas.0 45.0° 2.00 km 3.50 Third leg km
The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius R of the Earth, 8, and h. (1+ )² h\-2 8h = Suppose a 91.75 kg hiker has ascended to a height of 1.880 x 10³ m above sea level in the process of climbing Mt. Washington. By what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use g 9.807 m/s? and R : 6.371 × 106 m. Pay careful attention to significant figure rules, and enter your answer as a positive value.
1.3). Jisoo drives 5.25 km, 25° W of N, then 2.50 km, 65° S of W, and then 1.50 km, 345° w/ respect to the +x-axis. Find the magnitude and direction of the resultant displacement, using the method of components.
In astronomy, a unit of distance is used called the Astronomical Unit. It is the average distance from the Earth to the Sun. If 1 au is 1.49 × 1011 m, and a light-year is the distance it takes light to travel in 1 year elapsed time (light travels at a speed of c = 3.0 x 108 ), then how many au is it to get to Alpha Centauri (our closest star with a planetary system) if is 4.367 lightyears away?
You receive orders to sail due east for 24.6 mi to put your salvage ship directly over a sunken pirate ship. However, when your divers probe the ocean floor at that location and find no evidence of a ship, you radio back to your source of information, only to discover that the sailing distance was supposed to be 24.6 nautical miles, not regular miles. Use the Length table below to calculate how far horizontally you are from the pirate ship. Length cm МЕTER km in. ft mi 1 centimeter = 1 102 105 0.3937 3.281 x 102 6.214 × 106 1 METER = 100 1 103 39.37 3.281 6.214 x 104 1 kilometer = 10° 1000 1 3.937 x 104 3281 0.6214 1 inch =2.540 2.540 x 102 2..540 × 105 1 8.333 x 102 1.578 × 105 1 foot = 30.48 0.3048 3.048 × 104 12 1 1.894 x 10* 1 mile = 1.609 x 105 1609 1.609 6.336 x 104 5280 1 1 angström = 10-10 1 fermi = 1015 m 1 fathom = 6 ft 1 rod = 16.5 ft m 1 nautical mile = 1852 m 1 light-vear = 9.461 × 1012 km 1 Bohr radius = 5.292 x 101" m 1 mil = 10* in. = 1.151 miles = 6076 ft 1 parsec =…
v(t)A 7 0 1 2 3 45 6 1 Time (in hours) A particle starts out from the origin. Its velocity, in miles per hour, is given by: v(t) = Vi +t, where t is the number of hours since the particle left the origin. How far does the parti- cle travel during the second, third, and fourth hours (from t = 1 to t = 4)? Velocity (in miles per hour)
A spelunker is surveying a cave. She follows a passage 160 m straight west, then 230 m in a direction 45degree east of south, and then 280 m at 30 degree east of north. After a fourth unmeasured displacement, she finds herself back where she started.