What is the frictional force between the tires of a 2000 kg car and an asphalt road if the coefficient of friction is 1.2?
Look up the density of saltwater and the density of fresh water. The ship will sit higher in the liquid with greater density. The density of the water displaced times the volume of the water displaced is equal to the weight of the ship. A fluid with a greater density will requ...
Let the new point be (x1, y1, z1) Then, using the distance formula: ((x1-1)^2 + (y1-3)^2 + (z1-3)^2)^0.5 = 5 (x1-1)^2 + (y1-3)^2 + (z1-3)^2 = 25 Put all units in terms of one variable. I'm choosing x1 first: x1+2/3 = y1+1/2; y1 = x1 + 2/3 - 1/2 = x1 + 1/6 x1 + 2/3 = z1-3/2...
For the square with area 25 yd^2, x^2 = 25; x = 5; where x is the length of a side of the square. The perimeter of this square is 5*4 = 20 For the square with area 5 yd^2, y^2 = 5; y = 5^0.5; where y is the length of the side of this square. The perimeter of this square is 4*(...
pH = -log[H+] pOH = -log[OH-] and pH = pOH = 14 pOH = -log(0.01) = 2 2 + pH = 14; pH = 12
This is a perfect opportunity for you to develop the skill of reading your textbook and finding the necessary information. You will probably need to find functions that read and search data, separate data into components, and store data. Find these functions, and read through ...
Available is a 5 M/liter solution. You need 100 mL of a 0.91 M solution. In units of moles, you need 0.91 mol/liter * 0.1 liter = 0.091 moles of HNO3 0.091 moles / (5moles/liter) = 0.0182 liters of the 5 M solution. Take 0.0182 liters of the 5 M solution, and dilute to 100 mL...
V = 4/3*PI*r^3 dV/dt = 12 = 4*PI*r^2*dr/dt = 4*PI*1^2*dr/dt Solve for dr/dt, the rate at which the radius is growing
vboat + vcurrent = 23/3.6 vboat - vcurrent = 23/3 where vboat is the speed of the boat without a current, v is the speed of the current. The first equation models the boat traveling with the current; the second equation models the boat traveling against the current. Use algebr...
graph the function on your graphing calculator. The domain is the set of x values for which the function is defined; the range is the set of y values. Look at the graph and determine the set.