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agasfer [191]
3 years ago
11

The inner planets—mercury, venus, earth, and mars—are believed to have been formed by _____. answers

Physics
2 answers:
telo118 [61]3 years ago
7 0

<u>Answer:</u> These planets are formed by rocks.

<u>Explanation:</u>

There are 2 types of planets in Our Solar System:

1. <u>Inner planets:</u> There are 4 planets which are considered as inner planets, these are Mercury, Venus, Earth and Mars. These planets are small and have rocky surface. Their time of revolution around the Sun is less than the outer ones. These planets have maximum number of moon till 2.

2. <u>Outer planets:</u> There are 4 planets which are considered as outer planets, these are Jupiter, Saturn, Uranus and Neptune. These planets are large and gaseous planets. Their time of revolution around the Sun is more than the inner ones. These planets have many moon.

Hence, the inner planets are believed to be formed of rocks.

DaniilM [7]3 years ago
6 0
The first four planets Mercury, Venus, Earth and Mars, also called the inner planets are formed by rocks. While the <span>outer planets (Jupiter and beyond) are composed mostly of gas. 

Our answer to this question is Rocks.</span>
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A minibus drives with a constant speed of 76 miles per hour. How can I travel in 4 hours?
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A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
3 years ago
Q.3. The equivalent resistance across AB is:<br> (a)1<br> (c)2<br> (b)3<br> (d)4
sp2606 [1]

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

7 0
2 years ago
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