Concept explainers
(a)
The number of times terminal speed greater than for steel sphere that of the wooden sphere.
(a)
Answer to Problem 92P
The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.
Explanation of Solution
Write the expression for terminal velocity of the object.
Here,
The cross sectional area of the sphere is,
Here,
Rewrite the above equation for terminal velocity of wooden sphere.
Here,
Rewrite the equation (I) for terminal velocity of steel sphere.
Here,
Conclusion:
Solve the equation (III) and (IV).
Rewrite the above relation by using equation (2).
Substitute
Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.
(b)
The ratio of the terminal speed of the steel sphere to that of wooden sphere.
(b)
Answer to Problem 92P
The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.
Explanation of Solution
From part (a),
The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,
Conclusion:
Substitute
Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.
Want to see more full solutions like this?
Chapter 3 Solutions
College Physics, Volume 1
- An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.00 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 75.0 kg, and the mass of the sheet of metal is 245 kg. a. What is the velocity of the metal sheet relative to the honeycomb surface? b. What is the speed of the astronaut relative to the honeycomb surface?arrow_forwardIndividual masses on a line Sketch the following system on a number line and find the location of the center of mass. m1 = 8 kg located at x = 2 m; m2 = 4 kg located at x = -4 m;m3 = 1 kg located at x = 0 marrow_forwardOne of the first ion engines on a commercial satellite used Xenon as a propellant and could eject the ionized gas at a rate of 2.98 ✕ 10−6 kg/s with an exhaust speed of 3.04 ✕ 104 m/s. What instantaneous thrust (in N) could the engine provide? (Enter your answer to at least three significant figures. Enter the magnitude.)arrow_forward
- Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?arrow_forwardOne of the first ion engines on a commercial satellite used Xenon as a propellant and could eject the ionized gas at a rate of 2.97 ✕ 10−6 kg/s with an exhaust speed of 3.04 ✕ 104 m/s. What instantaneous thrust (in N) could the engine provide? (Enter your answer to at least three significant figures. Enter the magnitude.) Narrow_forwardFour objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.) m1 (kg) m2 (kg) m3 (kg) m4 (kg) 3.50 1.70 4.10 2.30 Four particles connected by light rods in the shape of a rectangle are shown. It lies on an x y-coordinate system with the center of the rectangle at the origin. An m2 particle lies in the first quadrant, an m1 particle lies in the second quadrant, an m4particle lies in the third quadrant and an m3 particle lies in the fourth quadrant. The rods that are parallel to the x-axis are of length 4.00 m. The rods that are parallel to the y-axis are of length 6.00 m. (a) Find the moment of inertia of the system about the x-axis.kg · m2(b) Find the moment of inertia of the system about the y-axis.kg · m2(c) Find the moment of inertia of the system about an axis through O and perpendicular to the page.kg · m2arrow_forward
- A thin wire is bent into the shape of a semicircle x2 + y2 = 4 x>0 If the linear density is a constant k find the mass and center of mass of the wire.arrow_forwardConsider this: Both scenario has same length of surface of 60 meters. A. A 50-kg object is located exactly 30 meters from both side of the surface. B A 75-kg object is located 20 meters from the left edge of the surface. Draw a free body diagram and write a short description about it.arrow_forwardAn astronaut of mass 180 kg including his suit and jet pack wants to acquire a velocity of 1.8 m/s to move back toward his space shuttle. Assuming the jet pack can eject gas with a velocity of 61 m/s, what mass of gas will need to be ejected? Express your answer to two significant figures and include the appropriate units. m=__________arrow_forward
- You go on a hot air balloon ride. The total mass of the balloon and the people riding in it is 1000 kg. At a height of 500 m, the balloon is rising straight up at a velocity of 10 m/s. The pilot drops 3 bags of sand out of the balloon basket. The new mass of the balloon and occupants is 850 kg. What is the velocity of the balloon after the sand is dumped? Group of answer choices 22.3 m/s 0 m/s 5.8 m/s 30 m/s 8.6 m/s 11.8 m/sarrow_forwardThree glasses are placed by a waiter on a light-weight tray. The first glass has a mass of M1 = 625 g and is located R1 = 13 cm from the center of the tray at an angle θ1 = 35 degrees above the positive x-axis. The second glass has a mass of M2 = 325 g and is located R2 = 24 cm from the center of the tray at an angle θ2 = 45 degrees below the positive x-axis. The third glass has a mass of M3 = 225 g and is located R3 = 17 cm from the center of the tray at an angle θ3 = 45 degrees above the negative x-axis. A fourth glass of mass M4 = 825 g is to be placed on the tray so that the center of mass is located at the center of the tray. Part (c) Write a symbolic equation for the vertical position from the central y-axis that the fourth glass must be placed so that the vertical center of mass of the four glasses is at the center of the tray. Part (d) Calculate the numeric value of the vertical position from the central y-axis of the fourth glass in cm. Part (e) Calculate the distance…arrow_forwardA particle of mass 2.4 kg is subject to a force that is always pointed towards the North but whose magnitude changes quadratically with time. Let the y-axis point towards the North. The magnitude of the force is given as F = 6t2, and has units of newtons Determine the change in velocity Δv, in meters per second, of the particle between t=0 and t=1.3s. Determine the change in y coordinate, in meters, of the particle Δ y betweent=o and t=1.3 if the initial velocity= 15.9 m/s and directed North, in the same direction as the force.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning