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3 years ago

Where in the cell can you find the location of inherited traits?

Physics

2 answers:

riadik2000 [5.3K]3 years ago

8 0

D.n.a is the answer

sashaice [31]3 years ago

5 0

Answer:

D.N.A

Explanation:

The d.n.a is where all the heriditary information is stored.

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In a Millikan experiment, a droplet of mass 4.7 x 10^-15 kg floats in an electric field of 3.20 x 104 N/C.

andrezito [222]

Answer:

a. The force of gravity on the droplet is approximately 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity is approximately -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is approximately -1.439375 × 10⁻¹⁸ C

d. There are approximately 9 excess electrons on the droplet

Explanation:

The parameters of the Millikan experiment are;

The mass of the droplet, m = 4.7 × 10⁻¹⁵ kg

The electric field in which the droplet floats, E = 3.20 × 10⁴ N/C

a. The force of gravity on the droplet, F = The weight of the droplet, W = m × g

Where;

g = The acceleration due to gravity ≈ 9.8 m/s²

W = 4.7 × 10⁻¹⁵ kg × 9.8 m/s² = 4.606 × 10⁻¹⁴ Newtons

∴ The force of gravity on the droplet = W = 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity, $F_v$ = -W = -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is given as follows;

$F_v$ = q·E

∴ The electric force that balances the force of gravity, $F_v$ = q·E = -4.606 × 10⁻¹⁴ N

q·E = -4.606 × 10⁻¹⁴ N

q = -4.606 × 10⁻¹⁴ N/E

∴ q = -4.606 × 10⁻¹⁴ N/(3.20 × 10⁴ N/C) ≈ -1.439375 × 10⁻¹⁸ C

The excess charge, q ≈ -1.439375 × 10⁻¹⁸ C

d. The charge of one electron, e = 1.602176634 × 10⁻¹⁹C

The number of excess electrons in the droplet, n, is given as follows;

n = 1.439375 × 10⁻¹⁸ C/(1.602176634 × 10⁻¹⁹C) = 8.98387212405 electrons

∴ n ≈ 9 electrons.

7 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.2589 N when separated by

Harlamova29_29 [7]

Answer:

$q_1 = 7.19 \times 10^{-6} C$

$q_2 = -1.0 \times 10^{-6} C$

Explanation:

Let the initial charge on the two spheres are

$q_1, -q_2$

now we know that the force between them is given as

$F = \frac{kq_1q_2}{r^2}$

$0.2589 = \frac{kq_1q_2}{0.5^2}$

$q_1q_2 = 7.19 \times 10^{-12}$

now when two spheres are connected then final charge on them is given as

$q = \frac{q_1 - q_2}{2}$

now the force between them is given as

$0.3456 = \frac{k(q_1 - q_2)^2}{4(0.5)^2}$

now we have

$q_1 - q_2 = 6.19 \times 10^{-6}$

So by solving above two equations we have

$q_1 = 7.19 \times 10^{-6} C$

$q_2 = -1.0 \times 10^{-6} C$

3 0

3 years ago

can someone please help me answer this in the next 15 minutes. ill give a brainiest to whoever can help me. tysm if u do!! :))

Nataly_w [17]

Answer: a

Explanation:

4 0

3 years ago

Which of the following best describes what causes the pases of the moon.?

Alborosie

The correct answer is<span> The earth casts a shadow on the moon

The Earth casts a shadow on the moon and that's why we can't see the remaining parts. The parts that we can see are what we call the lunar phases.</span>

4 0

3 years ago