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2 years ago

The property of water that contributes to its ability to stick to certain surfaces is called

Physics

1 answer:

Gekata [30.6K]2 years ago

3 0

Answer: Adhesion

Explanation: Adhesion is the tendency of dissimilar particles or surfaces to cling to one another (cohesion refers to the tendency of similar or identical particles/surfaces to cling to one another). The forces that cause adhesion and cohesion can be divided into several types. This allows Particles in things like water to stick to surfaces

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Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

Archy [21]

Initial height of platform is 15.87m or 16m.

8 0

2 years ago

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog

DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0

2 years ago

An automobile with tires that have a radius of 0.260 m travels 76000 km before wearing them out. How many revolutions do the tir

Tomtit [17]

Answer:

Explanation:

circumference of the tyre = 2πr = 2 x 3.14 x 0.26 = 1.6328m

76000km = 76000000m

no of revolutions required

= 76000000/1.6328 = 46546 revolutions.

3 0

2 years ago

Name two types of evidence used to support the theory of evolution. Explain how scientists use each type of evidence to provide

bearhunter [10]

Well, one of the theories was that Charles darwin proved that theory evolution is real so that one theory.

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2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

2 years ago