Ask question

Ask question

All categories

1 day ago

7

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

below the surface. What was the total inward force on the 1.1-m-diameter pilot sphere in which Cameron sat?

2 answers:

serg [2.5K]1 day ago

8 0

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

Ostrovityanka [2.2K]1 day ago

4 0

The inward force acting is approximately 4.1 × 10⁸ N

$\texttt{ }$

Further explanation

Let’s revisit the Hydrostatic Pressure equation, expressed as follows:

$\boxed{ P = \rho g h}$

where:

P = hydrostatic pressure ( Pa )

ρ = fluid density ( kg/m³ )

g = acceleration due to gravity ( m/s² )

h = height of the liquid column ( m )

Now, let’s solve the problem!

$\texttt{ }$

Given:

depth of the ocean floor = h = 11 000 m

diameter of pilot sphere = d = 1.1 m

atmospheric pressure = Po = 10⁵ Pa

Required:

total inward force = F =?

Solution:

$F = P A$

$F = ( P_o + \rho g h ) A$

$F = ( P_o + \rho g h ) ( \pi d^2 )$ → Area of Sphere = π d²

$F = ( 10^5 + 1000 \times 9.8 \times 11000 ) ( \pi \times 1.1^2 )$

$F \approx 4.1 \times 10^8 \texttt{ Newton}$

$\texttt{ }$

Conclusion:

The inward force totals around 4.1 × 10⁸ N

$\texttt{ }$

Learn more

Buoyant Force:

Kinetic Energy:

Volume of Gas:

Impulse:
Gravity:

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

You might be interested in

the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce

Ostrovityanka [2208]

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

4 0

1 month ago

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

kicyunya [2264]

Answer:

The initial speed at which the baseball is thrown is 20.58 m/s

Explanation:

To determine the time taken to reach the peak height in projectile motion, the formula is:

t = (u sin θ)/g

Where u = initial speed of the baseball =?

θ = angle of launch above the horizontal

g = gravitational acceleration = 9.8 m/s²

1.05 = (u sin 30)/9.8

u can be calculated as (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0

14 days ago

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t

Yuliya22 [2438]

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

3. the field is oriented away from the axon

Explanation:

The calculation for the electric field is illustrated below:

E = k*|q|/r²

Where:

E = electric field; k = 8.98755*10⁹ N*m²/C²; r = distance separating the field being measured from the point charge = 0.05 m; q = point charge

For a length of 0.100 m of the axon, the value of q is calculated as:

q = (5.6*10¹¹)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Therefore:

q = (5.6*10¹¹)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Consequently:

E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

3 0

19 days ago

Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

Keith_Richards [2263]

a)

i) 120 s

ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c) $N=mg+m\frac{v_b^2}{R}$

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

$v_b=\frac{2\pi R}{T}$

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

$v_b=\frac{2\pi (30)}{120}=1.57 m/s$

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

$W=mg$

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

$F=ma$

where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.

c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, $W=mg$ , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force $m\frac{v_b^2}{R}$ , leading us to the equation:

$N-mg=m\frac{v_b^2}{R}$

where

$v_b$ represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

$N=mg+m\frac{v_b^2}{R}$

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

$N=mg+m\frac{v_b^2}{R}$

- The weight (gravitational force) of the book is

$W=mg$

Upon comparing these two equations, we conclude:

$N>W$

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.

4 0

1 month ago

Explain the challenges in developing an accurate rating system for earthquakes. What kinds of variables are there?

Yuliya22 [2438]

Answer:

The primary factors that influence the intensity of shaking during an earthquake include the depth of the quake, its distance from the fault line, the type of soil below, and specifics about the buildings—especially their height. We will focus on the last two (soil quality and structures) and their interaction.

6 0

21 day ago