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emmainna [20.7K]
3 years ago
9

Which trait would be inheritable?

Physics
2 answers:
dolphi86 [110]3 years ago
8 0
<span>B) eye color is an inheritable trait</span>
Marina CMI [18]3 years ago
5 0
B. eye color  because nothing else  is not true
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Someone help me please
riadik2000 [5.3K]

the answer is 25 newtons

6 0
2 years ago
Read 2 more answers
Will give correct answer brainliest
Sav [38]

Answer:

.....

Explanation:

i. sorry? i really hope this helps!

4 0
2 years ago
A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A
Bond [772]

Answer:

\mathbf{V_x = 3.25 \ cm/s}

\mathbf{V_y = 1.29\ cm/s}

Explanation:

Given that:

The radius of the table r = 16 cm  = 0.16 m

The angular velocity = 45 rpm

= 45 \times \dfrac{1}{60}(2 \pi)

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

t = \dfrac{r}{v}

t = \dfrac{0.16}{0.035}

t = 4.571s

So the angle made by the radius r  with the horizontal during the time the bug gets to the end is:

\theta = \omega t

\theta = 4.712 \times 4.571

\theta = 21.54^0

Now, the velocity components of the bug with respect to the table is:

V_x = Vcos \theta

V_x = 0.035 \times cos (21.54^0)

V_x = 0.0325 \ m/s

\text {V_x = 3.25 \ cm/s}\mathbf{V_x = 3.25 \ cm/s}

Also, for the vertical component of the velocity V_y

V_y = V sin \theta

V_y = 0.035 \times sin (21.54^0)

V_y = 0.0129\ m/s

\mathbf{V_y = 1.29\ cm/s}

6 0
2 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1
ahrayia [7]

Answer:

The temperature is  T  = 168.44 \ K

Explanation:

From the question ewe are told that

   The rate of heat transferred is    P  = 13.1 \ W

     The surface area is  A = 1.55 \ m^2

      The emissivity of its surface is  e = 0.287

Generally, the rate of heat transfer is mathematically represented as

           H  =  A e \sigma  T^{4}

=>         T  =  \sqrt[4]{\frac{P}{e* \sigma } }

where  \sigma is the Boltzmann constant with value  \sigma  = 5.67*10^{-8} \ W\cdot  m^{-2} \cdot  K^{-4}.

substituting value  

             T  =  \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }

            T  = 168.44 \ K

7 0
3 years ago
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