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coldgirl [10]
3 years ago
14

A hungry hawk was preying on a lizard who was running northwards to get away from the low-flying hawk. If the lizard can run 8m

across water in 4s, what is the lizard's velocity? (Speed: D/T)
Physics
1 answer:
Anastaziya [24]3 years ago
8 0

Answer:

2m/s²

Explanation:

velocity = displacement (distance in a specified direction /time

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If a person weighs 600N on earth, how much does he weigh on the moon
Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0
3 years ago
IF you are in Space and push a bowling what happens to you and the bowling ball?
Anastaziya [24]

Answer:

The bowling ball did not change size or shape- the only thing that changed was the amount of gravity that pulls on it. But the mass of the bowling ball would never change. A bowling ball with a mass of 12 pounds on earth will have the mass of 12 pounds on the moon! Mass is the amount of atoms that a space fills.

Explanation:

I hope this helps! :D

4 0
3 years ago
a spring is compressed 15 centimeters when a 8.5 kilogram weight is set upon it. how much power would be needed to stretch out t
o-na [289]

Answer:

159.38 Watts

Explanation:

Initially;

  • Mass on the spring is 8.5 kg
  • Therefore, compression force is 85 N
  • Compression distance is 15 cm or 0.15 m

But;

F = kx

where F is the force of compression, k is the spring constant and x is the compression distance.

Thus;

k = F/x

  = 85 N ÷0.15

  = 566.67 N/m

We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.

Power = Work done ÷ time

Work done is given by 0.5kx²

Therefore;

Power = 0.5kx²÷ t

          = (0.5×566.67 N/m × 1.5² ) ÷ 4 seconds

          = 159.38 Watts

Thus, the power needed is 159.38 watts

5 0
3 years ago
Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt
Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

= \frac{4}{3}\pi(6400+ 50)^3 -  \frac{4}{3}\pi (6400)
^3\\\\=  \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3

Hence the volume of atmosphere is 2.6\times 10^{19} m^3

(b)

Write the ideal gas equation as foll ows:

PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles

no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules


Hence the required molecules is 13.3\times10^{43} molecules


(c)

Write the ideal gas equation as follows:

PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles

no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules

Hence the required molecules in Caesar breath is 1.2\times 10^{23} molecules

(d)

Volume fraction in Caesar last breath is as follows:  

Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}

(e)

Since the volume capacity of the human body is 500 mL.

Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath

5 0
3 years ago
At the instant shown in the diagram, the car's centripetal acceleration is directed
serg [7]

Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

3 0
3 years ago
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