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Zolol [24]
3 years ago
7

A car travels for 10s at a steady speed of 20m/s along a straight road. The traffic lights ahead change to red, and the car slow

s down with a constant deceleration, so that it stops after an additional 8 seconds.
Use your graph to deduce how far the car has travelled during the 18 seconds described.
Physics
1 answer:
pentagon [3]3 years ago
5 0
Okay. Draw a graph. Just a square wuth 10 lines in it. And then you can do the problem on that. Its really simple. A car Travels for 10 S at a steady speed of 20 M/S along a Street Rd. The traffic lights ahead and change to read in the car shows down with a constant the deceleration, so that if it stops after in a dish and on eight seconds.

Use your graph yo deduce how far the car has traveled during the 18 seconds described.
18x8x20x10x88=
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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

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.

8 0
3 years ago
A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?
svlad2 [7]
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r-ruslan [8.4K]

Answer:

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in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

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Answer:_

Explanation:

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