If sound becomes louder, which one of theses is increasing?

(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t

Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

$P = P_{o} + \rho gh$

Where $P$ is the absolute pressure

$P_{o}$ is the atmospheric pressure

$\rho$ is the density

$g$ is the acceleration due to gravity (Take $g = 9.8 m/s^{2}$ )

h is the height

From the question

h = 30.0 m

$\rho$ = 1.00 × 10³ kg/m³ = 1000 kg/m³

$P_{o}$ = 101.3 kPa = 101300 Pa

Using the formula

$P = P_{o} + \rho gh$

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

5 0

3 years ago

1. The mass defect of iron-56 is 0.52875 amu. What is the energy equivalent of this mass?

Delvig [45]

Answer:

Hello! Your answer is BELOW

Explanation:

1.About 91.754% of all iron is iron-56. Of all nuclides, iron-56 has the lowest mass per nucleon. With 8.8 MeV binding energy per nucleon, iron-56 is one of the most tightly bound nuclei.

2.The atomic weight of lead is quite variable in nature because the three heaviest isotopes are the stable end-products of the radioactive decay of uranium (238U to 206Pb and 235U to 207Pb) and thorium (232Th to 208Pb).

3.Mass defect for uranium-238 is 3.983 × 10-25 kg.

4.Energy and Mass Are Relative

The equation E = mc^2 states that the amount of energy possessed by an object is equal to its mass multiplied by the square of the speed of light.

Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥

7 0

3 years ago

The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

Marianna [84]

Tangent to the pathh.

7 0

3 years ago

Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.

34kurt

Answer:

I believe the answer to be B

Explanation:

If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

5 0

3 years ago

Read 2 more answers

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$