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3 years ago

Rosalind franklin What was she famous for

Engineering

1 answer:

liq [111]3 years ago

5 0

Answer:

She was known for her work on X-ray diffraction images of DNA, which led to the discovery of the DNA double helix for which Francis Crick, James Watson, and Maurice Wilkins shared the Nobel Prize in Physiology or Medicine in 1962.

Explanation:

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A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m

Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

$F_c = \dfrac{mv^2}{r}$

$F_c = \dfrac{2000\times 20^2}{300}$

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67 = μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0

3 years ago

Please help!!

Inessa05 [86]

Answer:

I think that the answer is the Chinese

Explanation:

3 0

3 years ago

Read 2 more answers

A tax on the amount of money a person earns in a year is a

irina [24]

Answer: income tax

Explanation:

7 0

3 years ago

A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowr

hram777 [196]

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = $12\ \text{m}^3/\text{s}$

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}$

Froude number is given by

$Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5$

Since $F_r>1$ the flow is super critical.

Flow is critical when $Fr=1$

Depth is given by

$d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}$

The depth of the channel will be 1.2 m for critical flow.

4 0

2 years ago

In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample

frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

$T=(\frac{Cv}{d^{2} })t$

where Cv is the coefficient of consolidation

$(\frac{Cvt}{d^{2} })_{1}= (\frac{Cvt}{d^{2} })_{2}$

if Cv is constant, we have:

$(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2}) }=\frac{t2}{2^{2} }$

Clearing t2:

t2=32 min

3 0

3 years ago