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11111nata11111 [884]
1 year ago
13

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

sure is 1.0 × 10^5 Pa (1.0 × 10⁵ Pa). Calculate
Physics
1 answer:
timurjin [86]1 year ago
5 0

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

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A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of
qaws [65]

Answer:

\frac{E_1}{E_2}= 4

Explanation:

Capacitance C is given by

C= \frac{\epsilon_0A}{d}

A= area of capacitor cross section

d= distance

therefore,

C_1= \frac{\epsilon_0A_1}{d_1}

A_1= πR^2

d_1= d

C_2= \frac{\epsilon_0A_2}{d_2}

A_= π(2R)^2

d_2 = 2d

q= \frac{\epsilon_0A_1}{d_1}V_1

threfore

V_1= \frac{qd_1}{\epsilon_0A_1}

and

V_2= \frac{qd_2}{\epsilon_0A_2}

also we know that E= V/d

⇒\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}

⇒\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}

= A_1/A_2= \frac{4R^2}{R^2}=4

therefore,

\frac{E_1}{E_2}= 4

5 0
2 years ago
A nucleus whose mass is 3.499612×10^(−25) kg undergoes spontaneous alpha decay. The original nucleus disappears and there appear
Elanso [62]

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (6.5898 × 10⁻¹³) J

Explanation:

Old nucleus ---> New nucleus + alpha particle.

We will use the conservation of energy theorem for extremely small particles,

Total energy before split = total energy after split

That is,

Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)

Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)

Rest energy = mc² (Einstein)

Let Kinetic energy be k

Kinetic energy of original nucleus = k₀ = 0 J

Kinetic energy of new nucleus = kₙ

Kinetic energy of alpha particle = kₐ

Mass of original nucleus = m₀ = (3.499612 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.433132 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = c = (3.0 × 10⁸) m/s

Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)

Total energy of new nucleus = (mₙc²) + kₙ

Total energy of the alpha particle = (mₐc²) + kₐ

(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ

kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)

(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)

(kₙ + kₐ) = (3.0 × 10⁸)² [(3.499612 × 10⁻²⁵) - (3.433132 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (9.0 × 10¹⁶)(0.00007322 × 10⁻²⁵) = (6.5898 × 10⁻¹³) J

5 0
2 years ago
In a particular application involving airflow over a surface, the boundary layer temperature distribution may be approximated as
Anni [7]

Answer:

The Surface heat flux is -9205 W/m^2

Explanation:

 Explanation is in the following attachment    

8 0
2 years ago
What’s the equation that links with total input, efficiency, energy and useful output energy transfer
weqwewe [10]

Efficiency. The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. Efficiencies can be written as decimals like 0.33 or percentages 33%. To convert a efficiency expressed as a decimal to a percentage you need to multiply by 100.

8 0
2 years ago
Now assume that the boat is subject to a drag force fd due to water resistance. is the component of the total momentum of the sy
DochEvi [55]
Based on the given details with this question, I can say that the direction of motion is not conserved. This is because the boat is subjected to an external force because of water resistance. So, the answer for this question would be NO.
4 0
1 year ago
Read 2 more answers
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