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4 years ago

.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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3 years ago

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If a piece of an object is dropped down vertically is the moment of inertia gonna be 0? And why?

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The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.

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3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

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Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

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3 years ago

A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months

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2 years ago

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Aloiza [94]

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

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4 years ago

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

.In the Highscope curriculum _and _ are active partners in shaping the educational experience