What happens in each of freud's stages?

Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p

Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

$\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v$

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = $\frac{2\times r^2(d-\rho)}{9\times n}$

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = $[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}$ [/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= $\frac{100}{387\times10^{-5}}$

= 2.6 x 10⁴ s.

5 0

2 years ago

A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is

timofeeve [1]

Answer:

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Explanation:

(a)

Distance between the maximum and the minimum of the wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If 1.21 m contains 1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.

6 0

3 years ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

6 0

3 years ago

What will happen if you put a usb drive and a magnet too closely together?

kakasveta [241]

<span>If you put a magnet right next to a USB drive, depending on the strength of the magnet and the amount of steel, nickel or cobalt used in the construction of that particular model of USB drive, the drive would either adhere to, or not adhere to, the magnet. This would cause no other significant effects. The storage of data in solid state form (as in USB drives) is not magnetic in nature, so no deletion or any other damage of the stored data would occur.</span>

6 0

3 years ago

Read 2 more answers

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$