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Lyrx [107]
3 years ago
5

How ar intrusive igneous bodies classified

Physics
1 answer:
Dahasolnce [82]3 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 package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
3 years ago
A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.
Ilya [14]

Answer:

t = 0.192 \mu m

Explanation:

Path difference due to a transparent slab is given as

\Delta x = (\mu - 1) t

here we know that

\mu = 1.79

now total shift in the bright fringe is given as

Shift = \frac{D(\mu - 1)t}{d}

Also we know that the fringe width of maximum intensity is given as

\delta x = \frac{\lambda D}{d}

now we have

\frac{D}{d} = \frac{\delta x}{\lambda}

now the shift is given as

Shift = \frac{(\mu - 1) t \delta x}{\lambda}

given that the shift is

Shift = 0.37 \delta x

here we have

0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}

now plug in all values in it

0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}

t = 0.192 \times 10^{-6} m

t = 0.192 \mu m

3 0
3 years ago
Order these sounds from loudest (1)
xz_007 [3.2K]

Answer:

1. Airplane

2. Crowd at sporting

3. Event

4. Rainfall

5. Whisper

7 0
3 years ago
Read 2 more answers
The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un
Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

5 0
3 years ago
A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a
Naily [24]

Answer:

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

a=\dfrac{P}{M+Rt}-g

We know that

a=\dfrac{dV}{dt}

\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g

\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt

V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

This is the velocity of bucket at the instance when it become empty.

6 0
3 years ago
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