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Liono4ka [1.6K]
3 years ago
14

Which force is there between an ice skate and the ice​

Physics
2 answers:
andrew-mc [135]3 years ago
7 0
The answer is friction
Temka [501]3 years ago
7 0
Friction

At the same time, if there were no friction at all on ice, skating would be impossible, because it is the friction between the skate and the ice when a skater pushes off that starts the motion to begin with. And friction is also what allows a skater to ever come to a stop.
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A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
Thermodynamics
Akimi4 [234]

Answer:

E = 3.8 kJ

Explanation:

Given that,

The mass of the object, m = 10 g = 0.01 kg

The heat of fusion of  aluminum is 380 kJ/kg

We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:

E = mL

So,

E = 0.01 × 380

E = 3.8 kJ

So, the energy required to melt the mass is equal 3.8 kJ.

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wind energy - D

geothermal energy - E

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I think that the answer to that is true hope that helps
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