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shepuryov [24]
3 years ago
10

On planet Q the standard unit of volume is called guppi. Space travelers from Earth have determined that one liter = 38.2 guppie

s. How many guppies are in 150 liters?
Physics
1 answer:
ankoles [38]3 years ago
8 0

Answer:

5730 guppies

Explanation:

1 liter= 38.2 guppies

150 liters= 150×38.2

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Who was Georges Lemaitre and how did he contribute to the Big Bang theory?<br> (15 points)
Ivahew [28]
A Belgian Astronomer and Cosmotologist who formulated the modern Big Bang theory, which states that the universe began in a cataclysmic explosion of a small, primeval “super-atom”
7 0
3 years ago
C. You push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. If the sle
Svetllana [295]

Explanation:

7.7 m/s

1st determine you subject slade to you have a mass of 15 kg being subjected to a force of 180N

so.

180N/15kg=180(kg m)/s^2/15kg=12 ms^2

now determine how long u pushed it

d=0.5 A T^2

sutable the known values getting

2.5 m = 0.5 12 ms^2 T^2

2.5m= 0.6 ms^2 T^2

0.41667 s^2=T^2

0.645497224s=T

the velocity multiply the time by the acceleration giving

0.645497224 s 12 ms^2 =7.745966692 ms.

after rounding the 2 significant u get 7.7 ms

4 0
3 years ago
Which one of the 4 answers is it?<br><br><br>​
slavikrds [6]
5 n to the right I think Hope this helps
8 0
3 years ago
There is a simple pattern for determining if a binary number is odd. what is it and why does this pattern occur
frez [133]
The answer to this question would be the least number (right most number).

If you add an even number with even number, the result should be even. But if there is just one odd number added, it will be odd.
In binary, all the number is multiplied by 2^{x}. It is pretty clear that all the number should be even beside the first/lowest multiplier 2^{0}  which has potential to be 1. 
5 0
4 years ago
Read 2 more answers
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

5 0
3 years ago
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