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Fynjy0 [20]
3 years ago
5

A drag-racing car goes fro 0 to 300mph in 5 seconds. What is its average acceleration in g's?

Physics
1 answer:
Solnce55 [7]3 years ago
8 0
I think that the short answer to this question is to convert 300mph into metres per second, and then divide by 5 to get the average acceleration. Since 1g is, I think, 9.81 m/s^2 (nearly 10), dividing by 9.81 should give the number of gs in the acceleration.1 mile is 1760 yards which is 1760x3 feet which is 1760x3x12 inches.300 miles would be 300x1760x3x12 inches.1 inch is 2.54 cm.300 miles would be 2.54x300x1760x3x12 cm.300 miles would be 2.54x300x1760x3x12/100 m.Then do the number of seconds in an hour.I hope this helps ...
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3 years ago
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Create a list of at least five different shapes of signs. For each of the five items on your list, identify two specific signs.
Nitella [24]
Hey bud, Good to see you again!

Okay so lets get to work huh? :D

Circle - Railroad advance warnings, and some bus signs.
Octagon - Stop signs, and slow signs (for flooding on roads).
Triangle - People crossing, People working.
Square - parking, speed limit.
Diamond - Bicyclist, animal crossing.

Hope this helped. Have a great day!


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3 years ago
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Suppose you want to operate an ideal refrigerator with a cold temperature of -10.0ºC , and you would like it to have a coefficie
Arlecino [84]

Answer: temperature of the hot reservoir is 303.52K

Explanation:

Th=temperagture of hot reservoir

Tc=temperature of cold reservoir 263K

Eff=efficiency of the refrigerator

Cop=coefficient of performance

7

Eff=1/cop

Eff=1/7

Eff=0.142

Eff=1-263/th

0.142-1=-263/th

-0.858=-263/th

Th=303.52K

7 0
3 years ago
A 215-kg load is hung on a wire of length of 3.60 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2. What
Y_Kistochka [10]

Answer:

4.74 * 10^-3

Explanation:

6 0
3 years ago
On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A
Anvisha [2.4K]

From the law of atmosphere

N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}

Where

n_0 = constant and is number density where the height y = 0cm

n_V = Number density at height y=3cm

Kb = Boltzmann constant = 1.38*10^{-23}J/K

T=20K

m = 10^{-19}kg

Re-arranging the equation to have the value of the gravity,

\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}

ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}

Since it is 30% of value above surface, therefore N_v = 0.3n_0

ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}

g = -\frac{KbT ln(0.3)}{my}

g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}

g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}

g = 1.104*10^{-1}m/s^2

g = 0.1m/s^2

Therefore the correct answer is C.

4 0
4 years ago
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