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Lyrx [107]
4 years ago
5

How ar intrusive igneous bodies classified

Physics
1 answer:
Dahasolnce [82]4 years ago
3 0
1. D
Intrusive igneous bodies are rock bodies that are formed due to the flow of magma. In classifying igneous bodies, their size, shape, and relationship(s) with the surrounding rock layers are all considered. 

2. B.
A batholith is the largest type of igneous body and it extends to unknown depths within the Earth's crust. These bodies are the size of mountains and are composed mainly of intermediate rock types like granite. 

3. B
This igneous body is formed when magma intrudes into rock fractures and then crystalizes there. Through this crystallization, the newly formed rock cuts through the existing rock layers.
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A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw
ZanzabumX [31]

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

W = 2N

so we have

m = \frac{W}{g}

m = 0.204 kg

now tension force at the top is given as

T_{top} = 7 N

T_{bottom} = 15 N

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

F_{net} = T + W

F_{net} = 7 + 2 = 9 N

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

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3 years ago
What is the difference between an atom in the ground state and an atom in an excited state
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The atom in an excited state has more energy and is less stable than the atom in the ground state.
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4 years ago
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How many normal modes of oscillation or natural frequencies does each if the following have: (
Vadim26 [7]
<span>Each of these systems has exactly one degree of freedom and hence only one natural frequency obtained by solving the differential equation describing the respective motions. For the case of the simple pendulum of length L the governing differential equation is d^2x/dt^2 = - gx/L with the natural frequency f = 1/(2π) √(g/L). For the mass-spring system the governing differential equation is m d^2x/dt^2 = - kx (k is the spring constant) with the natural frequency ω = √(k/m). Note that the normal modes are also called resonant modes; the Wikipedia article below solves the problem for a system of two masses and two springs to obtain two normal modes of oscillation.</span>
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umka21 [38]

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8 0
3 years ago
A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes
Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0
4 years ago
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