Calculate the number of centimeters (cm) in 1.0 yard (yd). Some Common Equalities
Quantity Metric (SI) U. S Metric-U. S.
Length Ikm = 1000 m 1ft = 12 m 2.54 cm = l in (exact)
Im = 1000 mm 1 yd = 3 ft I m = 39.4 in
1cm = 10 mm 1 mi = 5280 ft 1 km = 0.621 mi
Volume IL = 1000 ml. 1 qt = 4 cups 946 ml = 1 qt
1dL = 100 ml 1 qt = 2 pt IL = 1.06 qt
ImL = 1cm^2 1 gal = 4 pt
Mass 1kg = 1000 1 Ib = 16 oz I kg = 2.20 Ib
1g = 1000 mg 454 g = 1 Ib
Time 1h = 60 min 1h = 60 min
1min = 60s 1min = 60s
a. 1.0 yd = 10 cm.
b. 1.0 yd = 91 cm.
c. 1.0 yd = 7.6 cm.
d. 1.0 yd = 36 cm.

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

Hey student studies gravity using objects that have the same mass which two objects have the greatest gravitational force acting between them a. 100kg 1.0m 100kg b. 100kg 2.0m 100kg c. 100kg 2.0m 100kg big d. 100kg big 3.0m 100kg big

Positrons are spontaneously emitted from the nuclei of