85 mi/h x 1 km/.621 mi x 60 min/1 h x 60 sec/1 min = 492,753 km/sec (490,000 km/sec if you address the concern of significant figures)
PE = m•g•h, where m is mass in kg, g is 9.8m/s2, and h is height in meters. Solve for h.h = PE/m•g = 175 Joules/(0.36kg)(9.8m/s2) = 49.6m = 5.0 x 101m rounded to 2 significant figures
The capacity of a cylinder would be its volume. V=(pi)(r^2)(h), where V = volume in cubic units, pi = 3.14159 rounded to six significant figures (you can round to a different value; ask your teacher), r is radius, and h is height.Suppose you want to find the volume of a 15-cm high cylinder with a diameter of 8.00 cm. The radius is half the diameter, or 4.00 cm. V = (3.14159) x (4.00 cm)^2 x (15 cm) = 754 cm^3 (rounded to three significant figures)
Capital H has 2 lower case h has none
4: h/h, h/t, t/h & t/t
1,079,252,848 km/h or 299,792,457.8m/s
70km/h x (1000m/1km) x (1h/3600s) = 19.4m/s (without significant figures)
1mi = 5280ft 1h = 3600s 3.4mi/h x 5280ft/mi x 1h/3600s = 5.0ft/s (rounded to two significant figures)
Kph is km/h and mph is mi/h. 1 mi = 1.61km.75 km/h x 1mi/1.61km = 47 mi/h or 47 mph (rounded to two significant figures due to 75 kph).
d = 542mi s = 60mi/h s = d/t t = d/s t = (542mi)/(60mi/h = 9.03333h, or 9h, rounded to correct number of significant figures.
h
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h
AnswerLet x and y be any real numbers:log x = yx = log inv (y) = 10^yExample:pH =13.22 = -log [H+]log [H+] = -13.22[H+] = inv log (-13.22) = 10^(-13.22)[H+] = 6.0 x 10-14 MFINDING ANTILOGARITHMS using a calculator (also called Inverse Logarithm)Sometimes we know the logarithm (or ln) of a number and must work backwards to find the number itself. This is called finding the antilogarithm or inverse logarithm of the number. To do this using most simple scientific calculators,enter the number,press the inverse (inv) or shift button, thenpress the log (or ln) button. It might also be labeled the 10x (or ex) button.Example 5: log x = 4.203; so, x = inverse log of 4.203 = 15958.79147..... (too many significant figures)There are three significant figures in the mantissa of the log, so the number has 3 significant figures. The answer to the correct number of significant figures is 1.60 x 104.Example 6: log x = -15.3;so, x = inv log (-15.3) = 5.011872336... x 10-16 = 5 x 10-16 (1 significant figure)Natural logarithms work in the same way:Example 7: ln x = 2.56; so, x = inv ln (2.56) = 12.93581732... = 13 (2 sig. fig.)Application to pH problems:pH = -log (hydrogen ion concentration) = -log [H+] Example 8: What is the concentration of the hydrogen ion concentration in an aqueous solution with pH = 13.22? pH = -log [H+] = 13.22log [H+] = -13.22[H+] = inv log (-13.22)[H+] = 6.0 x 10-14 M (2 sig. fig.)
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Hexagon, a six-sided figure.