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3 years ago

Give an example of an artificial satellite.

1 answer:

4 0

The Hubble is a satellite. It orbits the Earth very quickly, completing one revolution every 97 minutes. This means the Hubble moves a speed of about 5 miles per second, or 18,000 miles per hour! When the Hubble observes distant stars and galaxies, it must stay pointed in the same direction for hours at a time. In addition, it must remain stable while fighting the effects of the Earth’s gravity and the solar wind.

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What are negative impacts of trenchwarfare

Mrac [35]

The main big one would be disease and infections. Trench fever, trench foot and trench mouth were common diseases people got when involved in trench warfare. Trench foot happens when the feet are exposed for long periods of time to cold water and mud. Trench fever is caused when body ice transfers from one place to another. Overcrowding conditions make this prone to happen. It causes headaches, sore muscles, skin lesions on the chest and backs as well as fever.

I hope this helps :)

4 0

3 years ago

Bacteria is being grown in a petri dish. The number bacteria B after t hours can be modeled by B(t) = 150e^0.585t. What is the i

Elina [12.6K]

The initial number of bacteria present is equal to 270 bacteria.
The number of bacteria that will be in the petty dish after 12 hours is 167818 bacteria.

<h3>What is an exponential function?</h3>

An exponential function can be defined as a mathematical function whose values are generated by a constant that is raised to the power of the argument. Mathematically, an exponential function is represented by this formula:

f(x) = eˣ

<h3>How to calculate the initial number of bacteria?</h3>

B(t) = 150e^0.585t

When t = 1, we have:

B(1) = 150e^0.585(1)

B(1) = 150e^0.585

B(1) = 269.25 ≈ 270 bacteria.

<h3>How many bacteria will be in the petty dish after 12 hours?</h3>

B(12) = 150e^0.585(12)

B(12) = 150e^7.02

B(12) = 167817.99 ≈ 167818 bacteria.

Read more on exponential functions here: brainly.com/question/27866047

#SPJ1

5 0

1 year ago

The process of desalination removes salt from seawater using condensation methods or filtration. Which of the following is least

lesantik [10]

Answer:

First is the high cost, because it takes a lot of increasingly expensive energy to remove salt from seawater. A second problem is that pumping large volumes of seawater through pipes and using chemicals to sterilize the water and keep down algae growth kills many marine organisms and also requires large inputs of energy (and thus money) to run the pumps. A third problem is that desalination produces huge quantities of salty wastewater that must go somewhere.

Explanation: hope this helped

7 0

3 years ago

You must __________ all emergency vehicles using a siren or flashing lights. A. honk to make your position known to B. follow C.

Tema [17]

The answer is C. You must always yield right-of-way to emergency vehicles using a siren and flashing lights.

Hope this helps!!

3 0

3 years ago

The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded

Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

$\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx$

Therefore;

$A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}$

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, $A_{1 shaded}$ = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

$A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526$

The shaded area, $A_{2 shaded}$ = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = $A_{1 shaded}$ + $A_{2 shaded}$

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0

3 years ago