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3 years ago

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Describe the basic internal structure of Mercury.

1 answer:

oksano4ka [1.4K]3 years ago

8 0

Answer:

has a radius of 2439 km (1524 mi), and the metallic iron-nickel core is believed to make up about 75% of this distance. Measurements of the planet's magnetic field made by Mariner 10 as it flew by the planet indicates that this core is likely to be hot and fluid.

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blsea [12.9K]

Answer:

True

Explanation:

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2 years ago

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What is meant by heat energys

ad-work [718]

Answer:

heat energy is the form of energy produced by heat

when we burn heat a type of enery is came

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2 years ago

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t

Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

$B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}$

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

$r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\$

<u>r = 0.05 m = 5 cm</u>

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3 years ago

A(n) ___ keeps a reaction from occurring by increasing the activation energy required for a reaction to occur. A. inhibitor B. c

skelet666 [1.2K]

Due to requirements i am typing this but the ans is a

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3 years ago

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You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

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3 years ago