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aivan3 [116]
3 years ago
13

The best place to look for crabs is between the high tide and low tide zones (the tidal flat). knowing this, would you expect to

find more crabs during a full moon/new moon or during a quarter moon stage? explain why.
Physics
1 answer:
disa [49]3 years ago
4 0
The best answer for this question is generally the best period to look for crabs is for the duration of the full moon, this is as a result of most invertebrates being nocturnal and being more effortlessly spotted and being fascinated to light sources.
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Compared to a plane gravitational pull on the ground to 7 miles in sky is what
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When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.
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Describe how all the air in the oven becomes hot?
vladimir1956 [14]
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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.

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QUESTION 1<br> 67.2 kilometers = how many decameters
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C cost (I need points)

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