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Slav-nsk [51]
4 years ago
6

.In the Highscope curriculum _____and _____ are active partners in shaping the educational experience

Physics
1 answer:
matrenka [14]4 years ago
8 0

Answer:

D. Community members, teachers

Explanation:

In the HighScope Curriculum teachers work in collaboration with family members, thus encouraging greater learning in students. They do this by providing information about the curriculum, inviting family members to participate in the activities carried out in the classroom, workshops for parents are also held. This allows discussing children's progress and sharing ideas to extend classroom learning to home.

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What is a greater force 10 Newtons or 5lbs?
Alexxandr [17]
5lbs is greater.
Hope that helps!
7 0
3 years ago
Read 2 more answers
A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s
sergey [27]

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0
3 years ago
The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?
NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

Hence, this is the required solution.

3 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
3 years ago
What is the correct answer?
damaskus [11]

Answer:

96 million of dollars

Explanation:

p=-6x^3+72x

p=-6(2)^3+72(2)

p=-6(8)+144

p=-48+144

p=96

8 0
3 years ago
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