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3 years ago

In 1990, the Human Genome Project began with the stated goal to ______.

2 answers:

8 0

The Human Genome Project was a large-scale international cooperation running from 1993 to 2004; it was performated in many countries, such as US, Canada, China and France.

Such a big project had huge goals: and in fact its most ambitious goal was to map <span>the entire human genome</span>

galina1969 [7]3 years ago

7 0

As far as I remember, in 1990, the Human Genome Project began with the stated goal to l<span>ocate the specific genes that cause given diseases.</span>

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A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the

kotykmax [81]

Answer:

M₂ = M then L₂ = L

M₂> M then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

∑ τ = 0

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

M L + M₁ 0 - m₂ L₂ = 0

M L - m₂ L₂ = 0

L₂ = $\frac{M}{M_{2}}$ L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

6 0

3 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If

wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by 850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0

3 years ago

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0

3 years ago

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge

Wittaler [7]

Answer:

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let's calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 - θ

θ ’= 180 - 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + 1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let's use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)

tea = 1,448 rad

to find the angle about the positive side of the + x axis

tea '= pi - 1,448

Tea = 1.6936 rad

6 0

3 years ago