Before answering this question, it is important to first understand the properties of sound waves:

From the question, it can be ascertained that the reflection of the sound wave has occurred.

As same as water wave, the sound wave obeys the law of reflection, which means the reflected angle is equal to the directed angle.

Sonar is transmitted into the sea via submarine.

When the sonar is detecting fish or rock, the sonar is reflected to the detector located on the ship.

Next, the detector will perform a calculation of the amount of depth of the sea or the distance of the fish or any object based on the amount of reflection time received.

Noted that,

\(\text {Sound speed}, v = \dfrac {\text {Distance}, s}{\text {Time}, t}\).

However, as the sound reflection requires a sound wave to move to the bottom of the sea or to the object and return to the ship, the distance will be multiplied by \(2\).

Source: noteforyou.com

The formula used to calculate sound reflections is:

\(\text {Sound speed}, v = \dfrac {2 (\text {Distance}, s)}{\text {Time}, t}\)

or more easily understood as:

\(\begin {aligned} \text {Sound speed}, v &= \dfrac {2 (\text {Distance}, D)}{\text {Time}, t}\\ &= \dfrac {2D}{t}. \end {aligned}\)

Rearrange the formula so that \(D\) can be left on the left side:

\(\begin {aligned} v &= \dfrac {2D}{t}\\ vt &= 2D \\ \dfrac {vt}{2}&= D \end {aligned}\)

where,

\(D = \dfrac {vt}{2}.\)

Enter all the information to find the depth of the sea and the distance between the object and the ship:

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(1.5)}{2} \\ &= 150 \space \text m \end {aligned}\)

while,

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(3)}{2} \\ &= 300 \space \text m. \end {aligned}\)

Therefore, it can be interpreted as the depth of the sea and the distance between the ship and the objects located at the bottom of the sea as follows:

Therefore, the thickness of the object is as follows:

\(\text {Thickness of the object}\\ = 300 - 150 \\ = 150 \space \text m.\)

The answer is option (C).

Sebelum menjawab soalan ini, adalah penting untuk memahami terlebih dahulu sifat-sifat gelombang bunyi:

Menerusi soalan, dapat dikenal pasti bahawa pantulan gelombang bunyi telah berlaku.

Sebagai mana gelombang air, gelombang bunyi turut mematuhi hukum pantulan iaitu sudut pantulan adalah sama dengan sudut tuju.

Sonar dipancarkan ke dalam laut menerusi bawah kapal.

Apabila sonar tersebut terkena pada ikan atau batuan keras, sonar tersebut akan dipantulkan semula ke alat pengesan yang terdapat di kapal.

Seterusnya, alat pengesan akan melakukan pengiraan jumlah kedalaman laut atau jarak ikan mahu pun objek berdasarkan tempoh masa pantulan yang diterima.

Perlu difahami bahawa:

\(\text {Laju bunyi}, v = \dfrac {\text {Jarak}, s}{\text {Masa}, t}\).

Namun begitu, disebabkan pantulan bunyi memerlukan gelombang bunyi untuk bergerak ke dasar laut atau ke arah objek dan kembali semula kepada kapal tersebut, maka, jarak akan didarabkan dengan \(2\).

Sumber: noteforyou.com

Formula yang digunakan bagi menghitung pantulan bunyi adalah:

\(\text {Laju bunyi}, v = \dfrac {2 (\text {Jarak}, s)}{\text {Masa}, t}\) ,

atau lebih mudah difahami sebagai:

\(\begin {aligned} \text {Laju bunyi}, v &= \dfrac {2 (\text {Jarak}, D)}{\text {Masa}, t}\\ &= \dfrac {2D}{t}. \end {aligned}\)

Susun semula formula agar hanya \(D\) sahaja yang ditinggalkan di sebelah kiri:

\(\begin {aligned} v &= \dfrac {2D}{t}\\ vt &= 2D \\ \dfrac {vt}{2}&= D \end {aligned}\)

yang mana,

\(D = \dfrac {vt}{2}.\)

Masukkan semua informasi bagi mencari kedalaman laut dan jarak antara objek dengan kapal:

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(1.5)}{2} \\ &= 150 \space \text m \end {aligned}\)

manakala,

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(3)}{2} \\ &= 300 \space \text m. \end {aligned}\)

Maka, dapat ditafsirkan kedalaman laut dan jarak antara kapal dengan objek yang berada di dasar laut seperti berikut:

Justeru, ketebalan objek adalah seperti berikut:

\(\text {Ketebalan objek}\\ = 300 - 150 \\ = 150 \space \text m.\)

Jawapan adalah pilihan (C).

Before answering this question, it is important to first understand the properties of sound waves:

From the question, it can be ascertained that the reflection of the sound wave has occurred.

As same as water wave, the sound wave obeys the law of reflection, which means the reflected angle is equal to the directed angle.

Sonar is transmitted into the sea via submarine.

When the sonar is detecting fish or rock, the sonar is reflected to the detector located on the ship.

Next, the detector will perform a calculation of the amount of depth of the sea or the distance of the fish or any object based on the amount of reflection time received.

Noted that,

\(\text {Sound speed}, v = \dfrac {\text {Distance}, s}{\text {Time}, t}\).

However, as the sound reflection requires a sound wave to move to the bottom of the sea or to the object and return to the ship, the distance will be multiplied by \(2\).

Source: noteforyou.com

The formula used to calculate sound reflections is:

\(\text {Sound speed}, v = \dfrac {2 (\text {Distance}, s)}{\text {Time}, t}\)

or more easily understood as:

\(\begin {aligned} \text {Sound speed}, v &= \dfrac {2 (\text {Distance}, D)}{\text {Time}, t}\\ &= \dfrac {2D}{t}. \end {aligned}\)

Rearrange the formula so that \(D\) can be left on the left side:

\(\begin {aligned} v &= \dfrac {2D}{t}\\ vt &= 2D \\ \dfrac {vt}{2}&= D \end {aligned}\)

where,

\(D = \dfrac {vt}{2}.\)

Enter all the information to find the depth of the sea and the distance between the object and the ship:

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(1.5)}{2} \\ &= 150 \space \text m \end {aligned}\)

while,

\(\begin {aligned} D &= \dfrac {vt}{2} \\ &= \dfrac {(200)(3)}{2} \\ &= 300 \space \text m. \end {aligned}\)

Therefore, it can be interpreted as the depth of the sea and the distance between the ship and the objects located at the bottom of the sea as follows:

Therefore, the thickness of the object is as follows:

\(\text {Thickness of the object}\\ = 300 - 150 \\ = 150 \space \text m.\)

The answer is option (C).