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1 year ago

Why does a block of plastic released under water come up to the surface of water.

1 answer:

7 0

A block of plastic released under water will come up to the surface of the water because the density of the plastic is less than the density of water.

<h3>What is density?</h3>

Density is the amount of mass a substance contains per its unit volume. It means the number of particles that can be packed into a unit space of that substance.

<h3>How does density determine whether an object will float or sink?</h3>

When an object is immersed in a liquid, it experiences two forces - gravitational force and an upward buoyant force called upthrust. If the density of the object is more than the density of the liquid, gravitational force will be greater than upthrust and the object will sink, but if the density of the object is less than the density of the liquid, upthrust will be greater than gravitational force and the object will float. That is why a block of plastic will come up to the surface when released under water, because it is less dense than water.

Learn more about density here:

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Do heavier bowling balls go faster than lighter ones?

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2 years ago

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A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around

kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

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2 years ago

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In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

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3 years ago

What is the volume of a cone with a height of 27 cm

JulijaS [17]

Explanation:

→ Volume of cone = πr² × h/3

Here,

Radius (r) = 13 cm
Height (h) = 27 cm

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

→ <u>Volume of cone = 4780.28 cm³</u>

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3 years ago

Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

andriy [413]

The answer I believe is False

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2 years ago