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4 years ago

Which of these substances would rise to the top if placed in a container with the rest?

Physics

1 answer:

Vladimir [108]4 years ago

4 0

What substances? Depends on their density, the lower density floats on top. For example, oil floats on top of water

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The process of sediments being compacted and cemented to form sedimentary rocks is called

crimeas [40]

Answer:

Lithification is the answer.

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3 years ago

A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the

crimeas [40]

Answer:

Explanation:

The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²

where Ф is the angle of inclination, R is the radius, k is the radius of gyration.

The potential energy of the system is given as ; PE = mgh

The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.

The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2

Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

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3 years ago

Why do liquids solidefy?

Setler79 [48]

energy extracted out of liquids an atoms are left to come closer arrange themselves shorter distance and then they solidify

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3 years ago

This exists in the space surrounding a charged particle and exerts a force on other charged particles.

Zinaida [17]

Gravitational field exists in the space surrounding a charged particle and exerts a force on other charged particles. Gravitational waves are ripples of waves travelling outward from the source. The more massive the orbit of two bodies, the more it emits gravitational wave. And everything around it that is near within the wave experiences a ‘pull’ toward the orbiting bodies.

4 0

3 years ago

Read 2 more answers

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

4 years ago