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RideAnS [48]
3 years ago
13

An effective early childhood curriculum can be​

Physics
1 answer:
kompoz [17]3 years ago
6 0
An effective early childhood curriculum can be learning how to read and write. Without learning how to read, we wouldn’t be able to communicate through this app, or through online school. We wouldn’t know directions, and we wouldn’t be able to enjoy stories. If we didn’t know how to write, I would not be able to help you figure out the answer to this problem, and your teacher wouldn’t be able to grade anything. Think about it, you ir entire education is based of being able to read and write.


Hope this helps :)
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1. The speaker uses needles from two
GuDViN [60]

Answer:

1. Georgia

2. Ten

3. Repeating

Explanation:

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3 years ago
Explain the process that causes dew to form on blades of grass. ​
AysviL [449]

Answer:

Condensation causes dew to form on blades of grass. This is because the water molecules react to the grass when it is heated, and the surroundings became cool. The molecules would bring to calm and turn to liquid.

Explanation:

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3 years ago
Read 2 more answers
A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct
Taya2010 [7]

Answer:

VB − VA = g tAB   &   (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s

v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s

\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s

\frac{h}{t}=\frac{3.8}{0.02}=190\ s

Hence, the equations VB − VA = g tAB   &   (VA + VB)/2 = h / tAB will be used

3 0
3 years ago
a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

8 0
3 years ago
Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel
coldgirl [10]

Answer: from the vertical, one should aim  86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² +  1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ  into equ 2

365.76(cos∅) ×  5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ =  3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0
3 years ago
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