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lubasha [3.4K]
2 years ago
5

Rocks in the Earth's mantle contain a higher percentage of _______ and _______ than do rocks in the Earth's crust.

Physics
2 answers:
Elenna [48]2 years ago
8 0
The Answer is C - iron and magnesium :)
kupik [55]2 years ago
5 0
The answer is c.......
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11.76 cm² into two significant figures and include the appropriate units
PilotLPTM [1.2K]
You will need to turn this into scientific notation, so:

1.176 × 10^1 cm^2

To two significant figures means to two digits so since 7 is greater than 5, you will need to round up.

Final answer is:

1.2×10^1 cm^2
8 0
3 years ago
A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.
harina [27]

I can not solve the problem if I do not have the mass.

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3 years ago
A concise definition of pair production
kogti [31]
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6 0
3 years ago
Hitungkan pecutan bagi blok di bawah: / Cal<br>(a)<br>m= 2 kg<br>F= 8.0 N​
ioda

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Acceleration = \frac {Net \; force}{mass}

Substituting into the formula, we have;

Acceleration = \frac {8}{2}

Acceleration = 4 m/s²

4 0
2 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
2 years ago
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