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3 years ago

Punnet square to show a cross between two short-haired cats

Physics

2 answers:

Bas_tet [7]3 years ago

6 0

Yes of coarse because 2=2 is 85

sergeinik [125]3 years ago

5 0

Are the cats heterozygous or homozygous dominant/recessive, is short hair dominant or recessive? cause if you don't know that then you can't do this problem

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The force between two charges, q, and 92, is F. If the distance between the

Sholpan [36]

Answer:

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0

3 years ago

A boy uses a force to keep a rock from falling. (He is carrying the rock.) Some forces, like this one, act only when two objects

MArishka [77]

Gravity is a non contact force , jump from a little high place (don't do that :P) , you wont just get stuck in the air you will fall down , this is gravity you dont need any contact

5 0

3 years ago

A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= $\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}$

= 1.72 x 10³ N.

8 0

3 years ago

Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ

Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

$F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}$

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

6 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of

Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$

m is mass of Singly charged uranium-238 ion, $m=3.95\times 10^{-25}\ kg$

q is charge

So,

$r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m$

So, the radius of their circular path is equal to 0.0548 m.

4 0

3 years ago