All categories

4 years ago

In general, how do the densities of a material in solid and liquid form compare?

2 answers:

mihalych1998 [28]4 years ago

6 0

Generally, a material in solid form is more dense than in liquid form. However, an exception to this rule is water, where the solid is less dense than the liquid (hence ice floats in water)

inna [77]4 years ago

4 0

The correct answer to the question is : The density of solid is more as compared to liquid.

EXPLANATION:

In case of solids, the molecules are closely aggregated to each other due to the strong inter molecular force of attraction. Hence, more masses are concentrated in a small volume. Hence, the density of solid is more.

In case of liquids, the molecules are not so closely aggregated to each other just like solids. Here, the inter molecular force of attraction is not so strong as compared to solid. Hence, density of liquid is less.

Hence, in general, the density of liquid is less as compared to solid.

You might be interested in

These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 305 nm radiation prov

astraxan [27]

Answer:

energy = 391.902 kJ /mol

Explanation:

given data

wavelength = 305 nm = 305 × $10^{-9}$ m

to find out

average energy

solution

we know speed of light is 3 × $10^{8}$ m/s

so we find frequency here first by speed of light formyla

speed = wavelength × frequency

3 × $10^{8}$ = 305 × $10^{-9}$ × frequency

frequency = 9.8360 × $10^{14}$ $s^{-1}$

so energy is

energy = hf

here h = 6.62 × $10^{-34}$ J-s

energy = 6.62 × $10^{-34}$ × 9.8360 × $10^{14}$

energy = 6.51 × $10^{-14}$ J

energy = 6.51 × $10^{-14}$ × $\frac{6.02*10^23}{1000}$ kJ/mol

energy = 391.902 kJ /mol

7 0

4 years ago

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t

labwork [276]

Answer:

<h3>(A) The width $x = 0.24 \times 10^{-3}$ m</h3><h3>(B) The new width is $1.32 \times 10^{-3}$ m</h3>

Explanation:

Given :

Focal length $f = 135 \times 10^{-3} m$

Maximum aperture $D = \frac{f}{4}$

Wavelength $\lambda = 550 \times 10^{-9}$ m

(A)

From rayleigh criterion,

$\theta = \frac{1.22 \lambda }{D}$

$\theta =\frac{ 1.22 \times 550 \times 10^{-9} }{33.75 \times 10^{-3} }$

$\theta = 1.98 \times 10^{-5}$ rad

From angle formula,

$x = R\theta$

Where $R =$ 12 m ( given in example )

$x = 12 \times 1.98 \times 10^{-5}$ m

$x = 23.76 \times 10^{-5}$

$x = 0.24 \times 10^{-3}$ m

(B)

We know that $\theta$ is proportional to the $x$ and inversely proportional to the $D$

so we write the new width, here $x$ is 5.5 times larger than above case

$x = 0.24 \times 10^{-3} \times \frac{22}{4}$

$x = 1.32 \times 10^{-3}$ m

6 0

3 years ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0

3 years ago

A spacecraft is in orbit around Mars. Suppose that it was previously observed to have speed 5 km/sec while at a distance r = 500

Lina20 [59]

Answer:

V' = 4.56Km/s

Explanation:

We can use the formula of Orbital Speed, who say,

$V = \sqrt{\frac{GM}{r}}$

The relative velocity is given by,

$V' = \sqrt{\frac{GM}{r'}}$

We need to find the relation between the two speeds,

$\frac{V'}{V} = \sqrt{\frac{GM}{r}*\frac{r'}{GM}}$

$\frac{V'}{V} = \sqrt{\frac{r}{r'}}$

$V' = V \sqrt{\frac{r}{r'}}$

Substituting,

$V' = (5)*\sqrt{\frac{5000}{6000}}$

$V' = 4.56Km/s$

6 0

3 years ago

a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses

sweet-ann [11.9K]

Answer:

Explanation:

Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .

Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²

Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²

Applying law of conservation of angular momentum

I₁ ω₁ = I₂ ω₂

Putting the values ,

3 x .75 = .27 x ω₂

ω₂ = 8.33 rad / s

New angular speed = 8.33 rad /s .

8 0

3 years ago