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3 years ago

8. How is the crystal size different for extrusive and intrusive igneous rocks?

1 answer:

eduard3 years ago

7 0

<span>Igneous rocks which form by the crystallization of magma at a depth within the Earth are called intrusive rocks. Intrusive rocks are characterized by large crystal sizes, i.e., their visual appearance shows individual crystals interlocked together to form the rock mass. hope that helped</span>

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This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.

hoa [83]

I believe the answer is c

4 0

3 years ago

Read 2 more answers

I need homework help

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

<span>6) On Earth, the gravity acceleration is </span> $g=9.81 m/s^2$ <span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span> $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$ <span>
</span>

5 0

3 years ago

Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius?

S_A_V [24]

<span>Jun 16, 2012 - Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius? –73°C 27°C 327°C 673°C.</span><span>
</span>

3 0

3 years ago

Read 2 more answers

A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrof

Vinil7 [7]

The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.

<h3>What is the specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol m₁ = 200-gram

The temperature of alcohol, T₁ = -6°C.

Mass of liquid sample of water m₂ = 400-gram

The temperature of the water, T₂= 20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

<h3 />

$\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C$

Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.

To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.

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2 years ago

Explain the role of good marketing in the success of business

Akimi4 [234]

Answer:

I am joker if you are mad x_________-..............,,,,,,........ €£

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2 years ago