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-BARSIC- [3]
3 years ago
6

Earth's layers formed as result of differentiation. During differentiation, different zones of elements form layers and today we

know these layers are made up of different elements. Certain elements sunk to the center while others collected near the surface. We might predict that the __________ elements, like iron and nickel, formed the core.
A) heavier
B) least dense
C) light weight
D) first crystallized
Physics
2 answers:
Afina-wow [57]3 years ago
6 0

The closest to the real answer is A, but the real answer is <u>more dense.</u>

elena-14-01-66 [18.8K]3 years ago
3 0
I think it's B) but I'm not positive
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Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ
const2013 [10]

Answer:

Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?

Explanation:

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2 years ago
Name the four layers of the atmosphere (in order starting at the bottom
Galina-37 [17]

Answer:

Troposphere, stratosphere, mesosphere and thermosphere. The next region is the exosphere, but that region is 500+ km from the Earth's surface.

6 0
2 years ago
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How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?
Rudiy27
The answer is the change in kinetic energy

K = 1/2 • mv^2

Change in K = (1/2 • 20 • 10^2) - (1/2 • 20 • 5^2)

Change in K = 1000 - 250 = 750J
5 0
3 years ago
Read 2 more answers
a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the
mihalych1998 [28]

Answer:

Part(a): The average acceleration is 1.83~m~s^{-2} during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if 'v_{i}', 'v_{f}' and 'a_{av}' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

a_{av} = \dfrac{v_{f} - v_{i}}{t}

where 't' is the time taken by the particle to achieve the velocity 'v_{f}' starting from initial velocity 'v_{i}'

Given in the problem, v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s.

So the average acceleration(a_{av}) during the first 6 s will be

a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.

7 0
2 years ago
Hooke’s Law Worksheet #1
dimaraw [331]

1. F = 0.25N

2. k = 80 N/m

3. x = 0.2m

4. a = 1m/s²

5. x = 42.6m

<u>Explanation:</u>

1. F = kx

  F = 2.5 X 0.1

  F = 0.25N

2. x = 0.25m

   F = 20 N

   F = kx

   20 = k X 0.25

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3. k = 3500 N/m

   F = 700 N

   x = ?

   700 = 3500 x

   x = 0.2m

4. m = 20kg

   k = 244 N/m

   x = 0.1m

   F = 20N

   a = ?

   F = ma

   20 = 20 a

    a = 1m/s²

5. k = 2.3 N/m

   m = 10kg

   x = ?

   F = kx

   10 X 9.8 = 2.3 x

    x = 42.6m

5 0
2 years ago
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