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- On Mars, a ball is rolling from a ground to a hill, and we know the height of the hill, Z, is 10 m, and the initial velocity ofthe ball on the ground and the final velocity on the hill are known: V1 = 20 m/s and V2 = 10 m/s, respectively. The mass ofthe ball, m, is 4 kg. The gravitational acceleration on Mars, gMars, is 3.7 m/s2. Please calculate 1) the change in kineticenergy, ΔKE, in J and 2) the change in potential energy, ΔPE, in J, of the ball for the rolling process.As you go above the Earth's surface, the acceleration due to gravity will decrease. Find the height, in (meters), above the Earth's surface where this value will be 1/150 g.Vmax In a laboratory experiment, a sphere of diameter 8.0 mm is released from rest at t = 0 at the surface of honey in a jar, and the sphere's downward speed v when it travels in the honey is found to be given by v = Vmax (1 - e-t/T), where 0.040 m/s and T = 0.50 s. (a) Obtain an expression for a(t). (b) Draw graphs for v(t) and a(t) for the time interval 0 to 2.0 s. (c) Obtain an expression for x(t), choosing the positive x axis as downward, and draw the graph for this func- tion. (d) Use your x(t) graph to determine the time interval needed for the sphere to reach the bottom of the container if the surface of the honey is 0.10 m above the bottom of the jar. ... =
- In physics, it is established that the acceleration due to gravity, g (in meters/sec2 ), at a height h meters above sea level is given by g(h) = 3.99 * 1014/(6.374 * 106 + h) 2 where 6.374 * 106 is the radius of Earth in meters. (a) What is the acceleration due to gravity at sea level? (b) The Willis Tower in Chicago, Illinois, is 443 meters tall. What is the acceleration due to gravity at the top of the Willis Tower? (c) The peak of Mount Everest is 8848 meters above sea level. What is the acceleration due to gravity on the peak of Mount Everest? (d) Find the horizontal asymptote of g(h). (e) Solve g1h2 = 0. How do you interpret your answer?a) A wind turbine with two or four hollow hemispherical cups connected to a pivot is commonly uscd to measure wind speed. Consider a wind turbine with four 8-cm- diameter cups with a center-to-center distance of 40 cm, as shown in Fig. PI1-35. The pivot is stuck as a result of some malfunction, and the cups stop rotating. For a wind speed of 15 m/s and air density of 1.25 kg/m3, determine the maximum torque this turbine applies on the pivot. A hemisphere at two different orientations for Re > 10 40 cm C04magnitude gravitational force (between a planet with mass 9.00 * 10 ^ 24 and moon, with 2.40 * 10 ^ 22 average distance between their canters 2.10 * 10 ^ 8* m b) ? is the moon's acceleration in m/s^ 2 ) toward the planet(Enter the magnitude. ) m/s^ 2 the planets acceleration (in m/s^ 2 ) the moon?
- The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104 light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the average speed of the Sun in its galactic orbit in m/s.The radius of the earth is R. At what distance above the earth's surface, in terms of R , is the acceleration due to gravity = 2.5 m/s2 ? At approximately (answer) × R above the earth’s surface.According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth's surface is mgR? F = (x + R)- where x = x(1) is the object's distance above the surface at time t, R is the earth's radius, and g is the acceler- ation due to gravity. Also, by Newton's Second Law, F = ma = m(dv/dt) and so dv m dt mgR? (x + R)? (a) Suppose a rocket is fired vertically upward with an initial velocity vo. Let h be the maximum height above the surface reached by the object. Show that 2gRh VR+ h [Hint: By the Chain Rule, m (dv/di) = mv (dv/dx).] (b) Calculate v. = lim--- Vo. This limit is called the escape velocity for the earth. (c) Use R = 3960 mi and g = 32 ft/s² to calculate v, in feet per second and in miles per second.
- Diwata-1 (aka PHL-Microstat-1) was launched to the International Space Station on March 23, 2016. It was the first Philippine microsatellite and the first satellite built and designed by Filipinos. This was released at a height of 400 kilometers above the Earth’s surface. Assuming a circular orbit, (a) how many hours does it take this satellite to make one orbit? (b) How fast (in m/s) is the Diwata-1 moving?In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day and he is blown along the following directions: 2.50 km and 45.0 north of west, then 4.70 km and 60.0 south of east, then 1.30 km and 25.0 south of west, then 5.10 km straight east, then 1.70 km and 5.00 east of north, then 7.20 km and 55.0 south of west, and finally 2.80 km and 10.0 north of east. Use a graphical method to find the castaway’s final position relative to the island.The “mean” orbital radius listed for astronomical objects orbiting the Sun is typically not an integrated average but is calculated such that it gives the correct period when applied to the equation for circular orbits. Given that, what is the mean orbital radius in terms of aphelion and perihelion?