All categories

4 years ago

If you can simply pour sand into a cup then why is it not a liquid?

2 answers:

7 0

If you pour sand into a cup(empty) why would it change?
If it had water in it then the sand <span>will sink to the bottom</span>, and not change forms.

sleet_krkn [62]4 years ago

6 0

If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.

You might be interested in

What is the difference between a tide and a wave.please answer.

Sedbober [7]

Answer:Tides are formed because of the interaction of the gravitational forces between the Earth, the moon, and the sun. ... Tides are made by the rising and falling sea levels with the action of gravity while waves are formed when several factors relating to the wind and water interact with each other

3 0

3 years ago

What the Ph measures??

Elenna [48]

PH measures the concentration of hydrogen ions ( $H^{+}$ ) in a solution. Knowing this, you can determine how acidic -or basic- your solution is.

5 0

3 years ago

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago

Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is give

Keith_Richards [23]

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

$W=F_G$

$mg = G \dfrac{mM}{R^2}$

which gives us an expression for the acceleration due to gravity <em>g</em> as

$g = G\dfrac{M}{R^2}$

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity $g_h$ at this height is

$mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}$

Simplifying this, we get

$g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g$

3 0

3 years ago

Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is

Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is $I = 0.822I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$

The distance from the screen is $D = 3.00m$

The intensity at the central maximum is $I_o$

The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

$I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

$sin \theta = \frac{z}{D}$

if the angle between the the light at z and the central maximum is small

Then $sin \theta = \theta$

Which implies that

$\theta = \frac{z}{D}$

substituting this into the equation for the intensity

$I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

We have that

$I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$

$=I_o [\frac{sin(0.760)}{0.760}] ^2$

$I = 0.822I_o$

4 0

3 years ago