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hodyreva [135]
2 years ago
15

Which of the following best describes a property of water?

Physics
1 answer:
NNADVOKAT [17]2 years ago
4 0

Answer:

C. weak cohesive forces exist between its molecules.

Explanation:

This is because water has less intermolecular forces than solids, but more than gases. Also their cohesive forces is low.

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The weakest type of friction that occurs between solid surfaces is _____ friction
Evgen [1.6K]

Answer:

Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.

Explanation:

4 0
2 years ago
Read 2 more answers
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
2 years ago
What will change the velocity of a periodic wave?
Fantom [35]
<span>it will be changed by changing the medium of the wave</span>
8 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
[10 POINTS ❗️❗️ and brainlist :)]
ludmilkaskok [199]
The answer to this is D
6 0
2 years ago
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