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a_sh-v [17]
4 years ago
7

If you know a car traveled 300 kilometers you can find it's ?

Physics
1 answer:
Artemon [7]4 years ago
5 0
If that's all the information you have, then the only thing you can infer
is that the number displayed on the car's odometer was 300 greater at
the end of the trip.
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Match each prefix to the correct multiple of 10. 1. centi- 10 2. kilo- 100 3. deka- 1000 4. milli- 1/100 5. hecto- 1/10 6. deci-
Mice21 [21]

Explanation :

There are some basic metric conversions.

Prefix                   Multiple of 10

centi                            \frac{1}{100}

Kilo                              1000

deka                              10

milli                              \frac{1}{1000}

hecto                            100

deci                              \frac{1}{10}

For example :

1cm=\frac{1}{100}m\\\\1Km=1000m\\\\1mm=\frac{1}{1000}m\\\\1Km=10\text{ hecto meter}\\\\1Km=100\text{ deka meter}


7 0
3 years ago
Read 2 more answers
How many valence electrons does each atom of arsenic (As) have? Arsenic is element 33. It is in period 4 and family 15 (5A or th
vodka [1.7K]

Answer:

It have 5 valence electrons

7 0
3 years ago
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If a car drives 10 miles NE and then due East for 10 minutes traveling 8
Nimfa-mama [501]
I think the answer is 28 miles
7 0
2 years ago
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Water waves approach an underwater "shelf" where the velocity changes from 2.8 m/s to 2.1 m/s. If the incident wave crests make
otez555 [7]

Answer:

24°

Explanation:

sin(34°)/sin(x)=v2/v1

x=arcsin(2,1*sin(34°)/2,8)=24°

7 0
3 years ago
A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
4 years ago
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