3 years ago

How u do it science

Physics

1 answer:

marusya05 [52]3 years ago

5 0

homozygous dominant:(BB) got 2 of them

heterozgous dimonant: (Bb) got 2 of them

homozygous Recessive: (none)

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How do I do this problem?

Aleks04 [339]

Use Charles Law: V1/T1 = V2/T2

0.30 m^3/27 C = V2/127 C

27V2 = 127 * 0.3

V2= 38.1/27 = 1.4 m^3

5 0

2 years ago

Mr Ndlovu works in centurion and works at Rosebank.It takes him 13minutes to get to work and he needs to be there at 08h15

VARVARA [1.3K]

Answer:

8:02 no less

Explanation:

8:15 - 13mins = 8:02

6 0

2 years ago

According to Ohm's law, a circuit with a high resistance _____. will have a low electric current will have a high electric curre

kakasveta [241]

Answer:

current

Explanation:

5 0

2 years ago

Read 2 more answers

A force of 20 N is exerted by an electric field on a test charge of 8.0 x 10² C at a point, P. What is the electric field streng

kenny6666 [7]

Answer:

the answer is equal to 1.6N/C

5 0

2 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

2 years ago