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maria [59]
2 years ago
10

Ice floats on water.​

Physics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

Yeah ice floats on water.

Observation

Example in those areas were ice is found like Antarctica ice is found on top of water.

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Help Plaeseeeeeeee!
vladimir2022 [97]

Answer:

golf all on grass covered hill in sunlight

Explanation:

someone pouring/spinning top has a lot of motion going on. a roller coaster has force acting on it which increases the motion.

3 0
2 years ago
Read 2 more answers
A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

6 0
3 years ago
Which microbes can be Eukaryotic? A. bacteria B. virus C. protists D. fungi
MissTica
Hey there,

Question : Which microbes can be Eukaryotic?

Answer : A, Bacteria

Hope this helps :D

<em>~Top♥</em>


3 0
3 years ago
Read 2 more answers
Jackson ran in a marathon for 3 hours. The marathon was 185 miles long. How fast was the runner running?
boyakko [2]

Answer:

hgt tdyrghgcderhgfe

Explanation:

rghttyhhhggggftttgf d's weyhvfswertfcdsettyhgftgdfvhgg fee e4y

4 0
2 years ago
Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
igomit [66]

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
3 years ago
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