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sukhopar [10]
3 years ago
9

A baseball pitcher throws a ball at 97.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)
Physics
1 answer:
mario62 [17]3 years ago
7 0

Answer:

Explanation:

1 mile=5280 ft

1 hour=3600 seconds

Changing speed from mi/h to ft/s

97 mi/h \times \frac {5280}{3600}=142.2666667 ft/s\approx 142.27 ft/s

Time of flight, t=\frac {distance}{speed}=\frac {60.5}{142.27}=0.425257732 s\approx 0.425 s

From fundamental kinematics equations

h=0.5gt^{2} where g is acceleration due to gravity whose value is taken as 32.2 ft/s2 and h is the distance

By substitution

h=0.5\times 32.2\times 0.425^{2}=2.9080625 m\approx 2.91 m

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Answer:

water

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How many Sig Figs (Significant Figures) are in each number?<br><br> 5070.0<br><br> 870.064080
nlexa [21]

∑ Hey, Lethality ⊃

Answer:

5070.0 has 5 significant figures

870.064080 has 9 significant figures

Explanation:

<u><em>Given:</em></u>

<em>How many Sig Figs (Significant Figures) are in each number?</em>

<em>5070.0</em>

<em>870.064080</em>

<u><em>Solution:</em></u>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>5070.0</em>

<em>5070.0 has 5 significant figures ( 5 , 0 , 7 , 0 , and 0 )</em>

<em>Number of significant figures: 5</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>870.064080</em>

<em>870.064080 has 9 significant figures ( 8, 7 ,0,0, 6,4,0,8 and 0 )</em>

<em>Number of significant figures: 9</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

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<u><em>xcookiex12</em></u>

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3 years ago
Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
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Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

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