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3 years ago

4. Which of the following would NOT be a major process in the formation of sedimentary

Physics

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

b. melting

Explanation:

it is made of sediments and that is not necessary

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Name your favorite food including desserts

AlexFokin [52]

Points? What do you mean

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3 years ago

Read 2 more answers

Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len

Mekhanik [1.2K]

Answer:

The difference in length for steel is 2.46 x 10⁻⁴ m

The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, $\gamma_{st.}$ = 3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, $\gamma_{in.}$ = 2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂ = L₁ + L₁αθ

L₂ - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = $\frac{\gamma}{3}$

Difference in length, for steel at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC$

ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC$

ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0

3 years ago

1pt The process by which rock minerals are changed by natural processes into new substances is known as:

vagabundo [1.1K]

The answer is A because you have to have erosion

8 0

3 years ago

A concert loudspeaker suspended high off the ground emits 33.0 W of sound power. A small microphone with a 0.600 cm2 area is 52.

Novay_Z [31]

Answer:

The sound intensity at the position of the microphone is $9.71\times10^{-4} W/m^{2}$

Explanation

Sound intensity is given by the formula

$I=\frac{P}{A}$

Where $I$ is the sound intensity, $P$ is the power and $A$ is the area.

Since the loudspeaker radiates sound in all directions, we have a spherical sound wave where the radius r is the distance of the microphone from the speaker.

∴ $A$ is given by $4\pi r^{2}$ where $r$ is the radius

From the question, $P$ = 33.0W, $r$ = 52.0m

$I=\frac{P}{A} = \frac{P}{4\pi r^{2} }$

$I = \frac{33.0}{4\pi \times (52.0)^{2} }$

∴ $I = 9.71\times10^{-4} W/m^{2}$

Hence, the sound intensity at the position of the microphone is 9.71 × 10⁻⁴ W/m²

7 0

3 years ago

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago