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NISA [10]
3 years ago
10

A fourth-grade teacher wants to ensure students understand the life cycle process of plants. The teacher decides to incorporate

the science-as-inquiry standard from the National Education Science Standards. How can the teacher achieve the specified learning outcome?
Physics
1 answer:
Annette [7]3 years ago
4 0

Answer:

There are several options that the teacher can use to incorporate the concept into students' understanding.

Explanation:

1. The students can draw all the plants that they know.

2. Children can be asked to bring the flowers to school so that they can identify the plants themselves.

3. The children can plat the flowers in makeshift pots and then take the best plants and transplant them in the garden or elsewhere.

4. The children can take occasional trips and observe and record any changes to the plants.

4. The teacher can ask the students to draw the flowers and emphasize on the productive parts like the stamens, leaves, pistils, stems.  

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In a stadium, fans stand up and sit down to produce a wave across the stadium. This type of wave where the material travels perp
puteri [66]
Correct option B

Transverse waves are those waves whose particles vibrate perpendicular to the direction of wave.

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3 0
3 years ago
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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
A child with a weight of 110 N swings on a playground swing attached to 2.00 m long chains. What is the gravitational potential
Ierofanga [76]

Answer

given,

Weight of the child = 110 N

length of the swing,L = 2 m

now, calculating the potential energy when the string is horizontal

  Potential energy = m g h

 now, h = L (1 - cos θ)   where θ is the angle made by the string with the vertical.

     PE = m g L (1 - cos θ)

   when rope is horizontal θ = 90°

     PE = 110 x 2 (1 - cos 90°)

    PE = 220 J

now, calculating potential energy when string made 25° with horizontal

PE = m g L (1 - cos θ)

   when rope is horizontal θ = 25°

     PE = 110 x 2 (1 - cos 25°)

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5 0
4 years ago
The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7
My name is Ann [436]

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

I=\dfrac{CB^2}{2\mu_0}

I=\dfrac{CE^2}{2\mu_0 C^2}

E=\sqrt{{2I\mu_0 C}}

E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}

E=930.84 N/C

Therefore answer is 930.84 N/C

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4 years ago
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It’s deeper in the orbit
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4 years ago
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