1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above

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

Question one please! This is 8th grade science btw.

yKpoI14uk [10]

Good Luck, NYAYAYAYAYa

8 0

3 years ago

A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permit

lara [203]

Answer:

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

$C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m$

The electric field between plates is given as:

E = V /d

$E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m$

3 0

3 years ago

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$