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3 years ago

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Fine grains of beach sand are assumed to be spheres of radius 49.8 µm. these grains are made of silicon dioxide which has a dens

ity of 2600 kg/m 3 . what is the mass of each grain of sand? answer in units of kg.

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer: $1.3\times10^{-9}kg$

Volume of each grain of sand:

$V=\frac{4}{3}\pi r^3$

radius of the spherical grain is given, $r=49.8\mu m=49.8\times10^{-6} m$

$\Rightarrow V=\frac{4}{3}\times3.14\times(49.8\times10^{-6})^3m^3=5.17\times10^{-13}m^3$

The grains are made up of silicon dioxide.

The density is equal to $D=2600 kg/m^3$

Then, the mass of each grain is given by:

$Mass=Density (D)\times Volume(V)\\ \Rightarrow M=2600 kg/m^3\times5.17\times10^{-13}m^3 =1.3\times10^{-9}kg$

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Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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Problem 6. A negatively charged particle is placed in a uniform electric field directed

Montano1993 [528]

<u>C</u><u>)</u><u> </u><u>South</u>

As we know that, north is considered as negative, and south as positive. Now, the charge on particle is negative; so, when we will release it, it will move towards the south.

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2 years ago

A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015

vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

$N=\dfrac{1}{e}N_0$

$N_0$ = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s$

Using exponential decay formula

$N=N_0e^{\dfrac{-bt}{m}}$

Final amplitude = Initial times decay

$\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s$

The time taken is 33.33 seconds

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On a certain map, every 3 cm represents 6 km on earth surface. Which ratio represents the scale of the map, which shows how the

damaskus [11]

Answer:

1 : 2 cm/km or $\frac{1}{2} \, \frac{cm}{km}$

One cm represents 2 km.

Explanation:

If 3 cm represents 6 km on the earth, the ratio to consider is: 3 : 6 cm/km which is the same as:

1 : 2 cm/km or $\frac{1}{2} \, \frac{cm}{km}$

One cm represents 2 km.

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Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant

evablogger [386]

(a) $2.56\cdot 10^{-5} C$

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

$F=(0.021)(1.1)=0.0231 N$

The electrostatic force between the two balloons can be also written as

$F=k\frac{Q^2}{r^2}$

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

$Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C$

(b) $1.6\cdot 10^{14}$ electrons

The magnitude of the charge of one electron is

$e=1.6\cdot 10^{-19}C$

While the magnitude of the charge on one balloon is

$Q=2.56\cdot 10^{-5} C$

This charge can be written as

$Q=Ne$

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

$N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}$

5 0

3 years ago