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USPshnik [31]
3 years ago
6

Compared to the image which drawing is the closest

Physics
2 answers:
LenaWriter [7]3 years ago
6 0

Answer:

The one on the left

Explanation:

Dmitrij [34]3 years ago
3 0

Answer:

The one on the left

Explanation:

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The sizes of most of the cells our bodies are about A.1micrometer
Alekssandra [29.7K]

Answer:

1Mm

Explanation:

i think this one. so it is wrong then also dont be angry with me because iam trying to help u

5 0
3 years ago
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Which state of matter has atoms that are spread out and bouncy?
NikAS [45]
The stage where atoms are spread out and bouncy is the gas stage.

7 0
3 years ago
A man weighs 1190 n on earth. what would he weigh on jupiter, where the free-fall acceleration is 25.9 m/s 2 ? the acceleration
Elden [556K]
The answer would be 3,145N. Using W=mg solve for the mass of the man on earth. Once you have the mass you can multiply it by the gravity of Jupiter giving you his weight in Newton’s on Jupiter.
8 0
4 years ago
A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0
3 years ago
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will
ExtremeBDS [4]

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = \frac{\Delta L}{L_o \ \Delta T}

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

5 0
3 years ago
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