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mezya [45]
3 years ago
15

The steps of mieosis

Physics
2 answers:
Grace [21]3 years ago
8 0

Answer:

here are six stages within each of the divisions, namely prophase, prometaphase, metaphase, anaphase, telophase and cytokinesis

Explanation:

lbvjy [14]3 years ago
4 0

Answer:

Since cell division occurs twice during meiosis, one starting cell can produce four gametes (eggs or sperm). In each round of division, cells go through four stages: prophase, metaphase, anaphase, and telophase.

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If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?
Alenkasestr [34]

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

F=\frac{G\times m_1\times m_2}{r^2}

G = gravitational constant

m_1, m_2 = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

F\propto \frac{1}{r^2}

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle  must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

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3 years ago
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5.0 m (1) (2) the slope of the line at any point in time

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Which of the following is not true about tectonic plates
AlladinOne [14]

Answer:

Scientists have identified about a dozen major and several minor tectonic plates

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Is eureka a better conductor than copper
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3 years ago
A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v
gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, d_{\frac{1}{2}}.

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

                                          d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, d_{\text{remain}} , in which the driver reduces the speed to 40km/hr is

                                             d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}.

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by  t_{\text{remian}}.

                                              t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}.

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

                                                     \text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}

Therefore, the average speed of the car is 50 km/hr.

8 0
2 years ago
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