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3 years ago

These are a musician favorite vegetable

Physics

2 answers:

Yanka [14]3 years ago

8 0

Beans.
As in the silly song "Beans, Beans, the magical fruit. The more you eat, the more you toot" lol

kicyunya [14]3 years ago

5 0

It is definitely an apple. I would know because i am a musician

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Is this answer correct?

alisha [4.7K]

Yes the answer is correct

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3 years ago

The average mass of a rhinoceros calf is 45 kg. What is its weight in newtons? Round your answer to the nearest whole number. 5

Thepotemich [5.8K]

Answer:

441 [N].

Explanation:

Weight=mass*g, where mass=45; g=9,8.

Weight=45*9.8=441 [N].

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3 years ago

Read 2 more answers

Which of the following hypotheses is both falsifiable and testable?

rosijanka [135]

B. If all plants are watered at least once a day then they will all grow at least 1 cm a day.

All plants grow at different rates.

Hope this helps! XD

5 0

3 years ago

Read 2 more answers

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

3 years ago

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

Ksju [112]

<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:

R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s

the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>

6 0

3 years ago