Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.

Answers

Answer 1

Answer:

B

Explanation:

P1/P1 = 40/20

=2


Related Questions

SI units are used for the scientific works,why?​

Answers

Answer:

SI is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion

which one is more powerful hydrogen bomb or atom bomb and why?​

Answers

Hydrogen bomb is more powerful than atom bomb

Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.

In the diagram, the crest of the wave is show by:
A
B
C
D

Answers

Answer:

D.

Explanation:

The crest of a wave refers to the highest point of a wave. This is illustrated by D.

What is science?Give two examples of living beings?

Answers

Answer:

the study of the past

Explanation:

dogs and cats

Find out other examples of bodies showing more than one type of motion Tabulate your findings.​

Answers

Answer:

down below

Explanation:

Image 1- wheels of train showing both translatory motion as well as rotatory motion.

Image 2- rotation of ball shows both rotatory motion as well as translatory motion.

Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)

Image 4- while cutting wood, the

carpenter's saw has both

translatory motion and oscillatory

motion, as it moves down while

oscillating.

a volcano that may erupt again at some time in the distant future is

Answers

The answer is a dormant volcano

If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?

a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.

Answers

Answer: They are equal in magnitude.

- The signs of the two changes in potential energy are opposite

Explanation:

When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.

Also, it should be noted that the signs of the two changes in potential energy are opposite.

Assume that the car on the left makes a quick turn to the left. According to inertia, your body will resist a change and still want to go in the original direction. In which direction with the passenger slide?

Answers

Answer:

to the right

Explanation:

if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.

Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained in them. What happens to the wires? What happens if the battery is replaced by an a-c source?

Answers

Answer:

(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.

(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.

hope it's help you ....!!!!!

plz mark as brain list and follow me #rishu...!!!!

Explanation:

Hope it will helps you lot!

g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?

Answers

Answer

Explanation

:giác mạc

Derive the dimension of coefficient of linear expansivity

Answers

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.

Q1. A metal rod is of length 64.576 cm at a temperature 90°C whereas the same metal rod has a length of 64.522 cm at a temperature 12°C. Calculate the coefficient of linear expansion.

A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

Answers

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

[tex] K = \frac{1}{2}mv^{2} [/tex]

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.

Answers

They made me do it I don’t even know what to say I’m so sorry

Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.

Answers

The result of the comparison of the appearance of Charon on Pluto and   times the Moon from Earth is that; Charon as seen from Pluto appears approximately 7 times larger than the Moon

Charon is directly overhead from the side of Pluto locked to the side of Charon

Charon appears at the horizon from the poles of the axis of rotation of Jupiter around Charon

The reason for arriving at the above solutions is as follows:

The given dimensions and distance from the Earth of the Moon are;

The diameter of the Moon, d = 3,474 km

The average distance of the Moon from the Earth, R = 384,000 km

Required:

The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth

Solution:

The distance of the Moon's travels in an orbit, C = 2·π·R

∴ C = 2 × π × 384,000 km

The angle subtended by the Moon, θ = d/C × 360°

∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°

Pluto's moon Charon, has the following parameters;

The diameter of the Charon, d₂ = 1,186 km

The average distance of the Charon from Pluto, R₂ = 19,600 km

Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂

∴ C₂ = 2 × π × 19,600 km

The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°

∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°

The angle subtended by Charon in Pluto's sky ≈ 3.415°

Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies

Required:

The appearance of Charon as seen from different locations on Pluto

Solution:  

Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon

Therefore;

Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and Charon

Learn more about Pluto's moon Charon here:

https://brainly.com/question/3920772

https://brainly.com/question/21590852

https://brainly.com/question/17177801

A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave

Answers

Answer:

a)  [tex]v=206.896m/s[/tex]

b)  [tex]L=6.749m[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=46Hz[/tex]

Length [tex]l=579m[/tex]

Total Mass [tex]T=4.3kg[/tex]

Tension [tex]T=3423[/tex]

a)

Generally the equation for velocity is mathematically given by

[tex]v=\sqrt{\frac{T}{\rho}}[/tex]

Where

[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]

Therefore

[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]

[tex]v=206.896m/s[/tex]

b)

Generally the equation for length of string is mathematically given by

[tex]L=\frac{3\lambda}{2}[/tex]

Where

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{206.89}{46}[/tex]

[tex]\lambda=4.498[/tex]

Therefore

[tex]L=\frac{3*4.498}{2}[/tex]

[tex]L=6.749m[/tex]

For example, we can take Water
In (A) Water has same mass and great volume
In (B) Water has same mass and lower volume
Will there be any change in its density then?

Answers

Answer:

yes there will be change in its density

Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?

Answers

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.

What is the potential energy of this pair of charges?

Answers

Answer:

5.4uC

Explanation:

True or false : conservation of energy gives a relationship between the speed of a falling object and the height from which it was dropped

Answers

Answer:

truee

Explanation:

Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?

Answers

Answer:

[tex]I_2=6.8A[/tex]

Explanation:

From the question we are told that:

Turns of inner coil [tex]N_1=120[/tex]

Radius of inner coil [tex]r_1=0.012m[/tex]

Current of  inner coil [tex]I_1=6.0A[/tex]

Turns of Outer coil [tex]N_2=150[/tex]

Radius of Outer coil [tex]r_2=0.017m[/tex]

Generally the equation for Magnetic Field is mathematically given by

[tex]B =\frac{ \mu N I}{2R}[/tex]

Therefore

Condition for the net Magnetic field to be zero

[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]

[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]

[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]

[tex]I_2=6.8A[/tex]

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B

Answers

Answer:

The speed at B is 16.18 ft/s .

Explanation:

Speed at A, u = 11 ft/s

Speed at C, v' = 18 ft/s

Time from A to C = 5 s

Time from B to C = 1.3 s

Let the speed of car at B is v.

Let the acceleration is a.

From A to B

Use first equation of motion

v = u + a t

18 = 11 + a x 5

a = 1.4 ft/s^2

Let the time from A to B is t' .

t' = 5 - 1.3 = 3.7 s

Use first equation of motion from A to B

v = 11 + 1.4 x 3.7 = 16.18 ft/s  

A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.

Answers

Answer:

A. Equals to that of the smaller sphere

B. 3 times less than that of the smaller sphere

Explanation:

(a) Equals to that of the smaller sphere

The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential

b) 3 times less than that of the smaller sphere

However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3

elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm​

Answers

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.

Answers

It will take a shorter amount of time for the cylinder to go down the plane down off the plane Because more pressure is applied one going up then going down there’s no pressure at all it’s the gravity is helping

Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do

Answers

Answer:

100 Joules

Explanation:

Applying,

W = mgh................... Equation 1

Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.

From the question,

Given: mg = 10 newtons, h = 10 meters.

Substitute these values into equation 1

W = 10×10

W = 100 Joules.

Hence the amount of workdone is 100 Joules

Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be

Answers

Answer:

a)  [tex]F_g=1.5*10^9Ibf[/tex]

b)  [tex]F_t=12490Ibf/ft^2[/tex]

     [tex]F_b=0[/tex]

Explanation:

From the question we are told that:

Height [tex]h=200ft[/tex]

Width [tex]w=1200ft[/tex]

a)

Generally the equation for Dam's Hydro static force is mathematically given by

[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]

Where

[tex]\rho=Density\ of\ water[/tex]

[tex]\rho=62.4Ibm/ft^3[/tex]

Therefore

[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]

[tex]F_g=1.5*10^9Ibf[/tex]

b)

Generally the equation for Dam's Force per unit area is mathematically given by

[tex]F=\rho*g*h[/tex]

For Top

[tex]F_t=\rho*g*h[/tex]

[tex]F_t=62.4*32.2*200[/tex]

[tex]F_t=12490Ibf/ft^2[/tex]

For bottom

[tex]Here \\H=0 zero[/tex]

Therefore

[tex]F_b=0[/tex]

The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

Hydrostatic force

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]

Force per unit area near the top

The force per unit area is the pressure exerted near the top of the dam.

[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]

The pressure near the bottom is zero, become the vertical height is zero.

Learn more about hydrostatic pressure here: https://brainly.com/question/11681616

A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.

Required:
What is the stiffness of a typical interatomic bond in the alloy

Answers

Answer: hello  some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness =  1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

Distance stretched (  Δl ) = 4 * 10^-3 m ( right value )

average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

        γ          = MgL / A Δl

                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

                    = 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness =   γ  * bond length

                           =  4756.36 * 2.3 * 10^-10  = 1.09 * 10^-6 N/m

What is hydroelectric power ?

Answer quickly..!

Answers

Answer:

It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.

Explanation:

One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.

A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?

Answers

Answer:

x = 68.3 m

Explanation:

tan 32.8 = x / 106

Electron A is fired horizontally with speed 1.00 Mm/s into a region where a vertical magnetic field exists. Electron B is fired along the same path with speed 2.00 Mm/s. (i) Which electron has a larger magnetic force exerted on it

Answers

B will have the greater force

Fc=MV2 /R=Fm

The A particle has less centipetal force and larger radius so larger curve

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Merchandise inventory, August 1 $ 96,610 Merchandise inventory, August 31 100,530 Purchases 254,660 Purchases returns and allowances 13,340 Purchases discounts 6,320 Freight in 4,070Cost of Merchandise Sold for August = ______.