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ira [324]
2 years ago
5

A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco

nds before he hears the "plunk" of the stone hitting the bottom of the shaft. Approximately how deep is the shaft? (Assume a gravitational acceleration of 9.8 m/s2 and no significant air resistance.)
A. 14 meters
B. 24 meters
C. 39 meters
D. 78 meters
Physics
1 answer:
Shalnov [3]2 years ago
3 0

The shaft is 78m approximately deep. The correct option is D which is 78 meters

<h3>What are Sound Waves ?</h3>

Sound waves are longitudinal. That is, the direction of the waves is parallel to the direction of its propagation of particles.

Given that a boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seconds before he hears the "plunk" of the stone hitting the bottom of the shaft.

  • The speed of the sound V = 330m/s
  • The time it takes the sound to reach the top = t

Speed V = distance / time

330 = h/t

Make t the subject of formula

t = h/330

As the stone is dropped, initial velocity = 0, Using the formula

h = 1/2gT²

But T = 4 - t

T = 4 approximately

Substitute all the parameters

h = 1/2 × 9.8 × (4)²

h = 4.9 × (16)

h = 78.4m

Therefore, the shaft is 78m approximately deep. The correct option is D which is 78 meters

Learn more about sound wave here: brainly.com/question/16093793

#SPJ1

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ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

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4 years ago
The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only
Sever21 [200]

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

\text{Initial Length} = L

\text{Initial surface Area} = 6L^2 (Surface of a Cube)

\text{Initial Volume} = L^3

The final measures

\text{Final Length} = L_f

\text{Final surface area} = 6L_f^2

\text{Final Volume} = L_f^3

Given,

\frac{(SA)_f}{(SA)_i} = 2

Now applying the same relation we have that

(\frac{L_f}{L_i})^2 = 2

\frac{L_f}{L_i} = \sqrt{2}

The relation with volume would be

\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3

\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3

\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})

\frac{(Volume)_f}{(Volume)_i} = 2.83

Volume of the cube change by a factor of 2.83

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Answer:

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Explanation:

45N/3kg=18

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A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change
olga55 [171]

Answer:

<em> The add mass = 5.465 kg</em>

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

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T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

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Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

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<em>Substituting into equation 4</em>

<em>m₂ = (2.07)²(0.5)/(1.18)²</em>

<em>m₂ = 4.285(1.392)</em>

<em>m₂ = 5.965 kg.</em>

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

<em>Thus the add mass = 5.465 kg</em>

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Answer:

A. both fields use the same technology

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