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2 years ago

Select the correct answer.

Physics

1 answer:

VLD [36.1K]2 years ago

8 0

Answer:O – H

Explanation:

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Which of the following effects is produced by the high surface tension of water? Which of the following effects is produced by t

Mashcka [7]

Answer:

A water strider can walk along the surface of earth due to the surface tension of water. 

Explanation:

Fluids have a tendency to shrink to minimum possible surface area and this is called surface tension. It usually occurs due to the greater force of cohesion between molecules of same substance when compared to adhesive force between molecules of different substances. Objects with greater densities can float along water surface due to the role played by surface tension. 

When insects walk along the water surface they are pulled down due to gravity. But the force of attraction between the legs of the insect and water molecules is minimal. Thus the surface tension would always tend to maintain the flatness of water overcoming the push by the legs of the strider.

When the insect’s weight pulls it down , the surface tension pushes it upwards overcoming this force of gravity. This is how water striders move along the surface of water.

7 0

2 years ago

Find the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y =

Simora [160]

The correct answer for the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y = 4x is 128/3 * π.

Volume of Solid of Revolution by Shell method is given by

V = 2π * integrate x(height) dx Here, height = 4x-x2

(1)& x-varies from x = 0 to x = 4 then from eqn(1) V = 2π * integrate x(4x - x ^ 2) dx from x = 0 to 4 = 2π * integrate (4x ^ 2 - x ^ 3) dx from x = 0 to 4

Basic Rule(1) ∫ x^n dx =x^ n+1/ n+1

V=2 π [4((x ^ 3)/3) - (x ^ 4)/4] 0 ^ 4 =2 π[ 4/3 x^ 3 - x^ 4/4 ] 0 ^ 4

V = 2π [4/3 * 4 ^ 3 - (4 ^ 4)/4} - 0]

V = 128/3 * π.

Learn more about area under curve here :-

brainly.com/question/15122151

#SPJ4

4 0

1 year ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

Two objects must be in contact for them to exert a force on each other. True or False

nevsk [136]

Answer:

False

Explanation:

This proposition is false because by example the sun exerts a force over the earth and them are not in contact

7 0

2 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

2 years ago