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blondinia [14]
2 years ago
15

A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h

alf-life of S is 3.9 s. When only Path 2 is active, the half-life of S is 104. s. Calculate the half-life of S when both pathways are active.
Physics
1 answer:
Misha Larkins [42]2 years ago
6 0

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

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Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?
galben [10]

Answer:

See explanation

Explanation:

First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.

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1 year ago
Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a
EleoNora [17]
Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ...  V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1;  -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1;  -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)
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1 year ago
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learn more:

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