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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 plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$

$M_t= 18,989.09 \ N-mm= 168.06 lb-in$

torque rating

$(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in$

4 0

3 years ago

From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye

Blababa [14]

Answer:

(a) Precipitation hardening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.

(4) The strength is developed by a heat treatment.

(b) Dispersion strengthening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(3) The hardening/strengthening effect is retained at elevated temperatures for this process.

(5) The strength is developed without a heat treatment.

7 0

3 years ago

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0

3 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

3 years ago

Omg help mr idk what to say ahhh

kap26 [50]

Explanation:

ответ на фото !!!!!!

7 0

3 years ago

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