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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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A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th
nadezda [96]
That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
6 0
2 years ago
Read 2 more answers
A man carries a load of mass 2.6kg from one end of a uniform pole 100cm long which has a mass 0.4kg. The pole rest on his should
miskamm [114]

Answer:

F = 39.2 N   (hand force) and    N = 68.6 N (shoulder force)

Explanation:

In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole

Let's set the reference system at the fit point of the shoulder

     ∑ τ = 0

We will assume that the counterclockwise turns are positive

    w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0

all distances are measured from the support of the man (x₀ = 0.60 m)

    F = (w₁ 0.60 + W 0.1) / 0.4

    F = (m₁ 0.6 + m 0.1) g / 0.4

let's calculate

    F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4

    F = 39.2 N

this is the force that the hand must exert to keep the system in balance

We apply the translational equilibrium condition

    -w₁ -W + N - F = 0

     N = w₁ + W + F

     N = (m₁ + m) g + F

let's calculate

     N = (2.6 + 0.4) 9.8 + 39.2

     N = 68.6 N

6 0
1 year ago
A friend of yours who has not taken an astronomy class looks at your textbook and really likes the picture of the Pleiades, a cl
Andrei [34K]

Answer:

<em>C. the blue colour of the Earth's sky</em>

<em></em>

Explanation:

The Pleiades is a cluster of sister stars that are among the closest star cluster to earth.

The reflection nebula of the Pleiades is due to the scattering of the blue light from the hot blue luminous stars that dominate the star cluster. Th blue light is scattered from dust molecules, thought to be predominantly carbon compound like diamond dusts, and other compounds like iron.

The blue colour of the Earth's sky is the closest terrestrial phenomenon to the reflection nebula. On a clear cloudless day, molecules in the air scatter the blue component of light more than the other component colours of white light, giving the sky its characteristic blue coluor.

The common characteristics of the luminous nebula and the Earth's blue sky is that they both have their light scattered by the presence of small particles.

8 0
2 years ago
A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds
horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

4 0
2 years ago
Read 2 more answers
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