Which is the only mechanism that can introduce new alleles in a gene pool? genetic drift random mating mutation

a sample contains 100 g of radioactive isotope. How much radioactive isotope will remain in the sample after 1 half-life?

kap26 [50]

Answer:

$\huge\boxed{50g}$

Definition:

Half-life- The time taken for half of the radioactive isotopes to decay.

Explanation:

How does radioactive decay work? Radioactive decay is a process by which unstable nuclei become more stable through the emission of alpha or beta particles or gamma rays.

Since a half-life is the time taken for half of the isotopes to decay, we can simply divide the initial mass of 100 grams by 2; this gives us 50 grams.

1) Divide 100g by 2.

$\frac{100g}{2}=50g$

8 0

2 years ago

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest

9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

$F = \frac{5978}{3}$

F = 1992.67 N

7 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

Explain why the gravitational acceleration of any object near earth is the same matter what was the mass of the object

Gre4nikov [31]

It takes greater force to accelerate an object that has more mass. But the gravitational force between the Earth and an object is greater when the object hass more mass. It works out just right to make any object with any mass accelerate at the same rate.

5 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$