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3 years ago

15

In 1949, dr. tony allison observed a high frequency of kenyans carrying the sickle cell allele in coastal areas and near lake vi

ctoria, but a lower frequency in the highlands. what did he hypothesize?

1 answer:

Lynna [10]3 years ago

8 0

Dr Tony Allision hypothesize that the sickle cell disease is connected with malaria. The disease carried by the mosquito is more common in coastal area and near Lake Victoria while it is less common in the highlands. He made a further hypothesis that sickle cell disease is common in areas where malaria is also common.

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A car accelerates uniformly in a straight line

julia-pushkina [17]

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

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3 years ago

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force

aniked [119]

Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

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3 years ago

An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(a). We need to calculate the induced emf

Using formula of emf

$E = Blv\sin\theta$

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(b). We need to calculate the power dissipated through the resistor

Using formula of power

$P=I^2 R$

Put the value into the formula

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

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3 years ago

Read 2 more answers

If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass

EleoNora [17]

Simply, apply the formula $E = \frac{1}{2}mv^{2}$ and insert the values of m = mass, v = velocity and E = Energy.
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3 years ago

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

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The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

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3 years ago