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stich3 [128]
2 years ago
9

Elyse is explaining to a friend that she will be conducting a scientific investigation. Her question is "How many leatherback se

a turtles will migrate to their home beach on the eastern coast of Australia this season?" Elyse states her hypothesis is "1,200 leatherback sea turtles will migrate to their home beach on the eastern coast of Australia this season." For her procedure, she and a team of scientists will be observing and counting the number of turtles that land on the beach. What is unusual about the description of her scientific investigation?
Physics
1 answer:
trapecia [35]2 years ago
8 0

<span>The scientific investigation has  five basic steps and one feedback step: (1) to make an observation (2) to ask a question (3) hypothesize (4) predict and (5) make a new hypothesis based on the results.  What is unusual about the description of Elyse’s scientific investigation is she did not follow the basic steps. She should have done observation first before asking a question and forming a hypothesis. </span>

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Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

6 0
2 years ago
Read 2 more answers
The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of
Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV    .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0
2 years ago
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
lord [1]

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

5 0
2 years ago
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a
Svetllana [295]

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 <em>u </em>s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 <em>u </em>s  = 75 <em>n</em> s

TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

8 0
2 years ago
When 0.1523 g of liquid pentane (CH) combusts in a bomb calorimeter, the temperature rises from 23.7C to 29.8 C. What is U for t
pashok25 [27]

Answer:

U for the reaction is 15,048.58 kJ/mol of pentane.

Explanation:

Quantity of heat required = heat capacity of bomb calorimeter × temperature rise = 5.23 kJ/C × (29.8 C - 23.7 C) = 5.23 kJ/C × 6.1 C = 31.903 kJ

Moles of pentane = mass/MW

Mass = 0.1523 g

MW of pentane (C5H12) = 72 g/mol

Moles of pentane = 0.1523/72 = 0.00212 mol

U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol

8 0
2 years ago
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