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love history [14]
3 years ago
9

Which of the following would be an example of positive eugenics?

Physics
2 answers:
mario62 [17]3 years ago
6 0
It would be A. performing DNA research to eliminate breast cancer.

Hope This Helps You!
Good Luck Studying :)
bija089 [108]3 years ago
3 0
Hello there.

<span>Which of the following would be an example of positive eugenics?

</span><span>A. Performing DNA research to eliminate breast cancer
</span>
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Yosemite falls, in california, has a total height of 73,900 cm. what is this height in meters?
Hunter-Best [27]
Yosemite falls has a total height of 73,900 cm it means it heights 739 meters
6 0
3 years ago
What is a heterogeneous mixture and what is a homogeneous mixture?
yuradex [85]
A heterogeneous mixture is a mixture that contains two or more distinct substances that you can see. You can see the different part if a heterogeneous mixture. An example of this is a salad. You can see all of the parts. A homogeneous mixture is a mixture that is uniform and you cannot see the different parts. It is still a mixture though. An example of that would be salt water. The water and salt are not chemically combined but you cannot see the salt AND water. It is just one solution.
5 0
3 years ago
Read 2 more answers
A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no
padilas [110]

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

F_N=mg\ cos(13).............(2)

The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}

\dfrac{F_N}{W_c}=cos(13)

\dfrac{F_N}{W_c}=0.97

or

\dfrac{F_N}{W_c}=\dfrac{97}{100}

Hence, this is the required solution.

5 0
3 years ago
The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth a
Nana76 [90]
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.

<span>The formula for force of attraction between any two bodies in the universe
F  =  GMm / r^2.      (Newton's Universal law of Gravitation).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11  Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22  kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
   = 3.8 x 10^8 m.

(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).


So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
  Punch it all up in your calculator.
 
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.

F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>




8 0
3 years ago
Read 2 more answers
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
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