A large rectangular tub is filled to a depth of 2.60 m with olive oil, which has density 915 kg/m3 . If the tub has length 5.00 m and width 3.00 m, calculate (a) the weight of the olive oil, (b) the force of air pressure on the sur- face of the oil, and (c) the pressure exerted upward by the bottom of the tub.

Answers

Answer 1
Answer:

Answer:

weight is 3.50 x 10^5 N

force is 1.52 * 10^6 N

pressure is 1.25 * 10^5 Pa

Explanation:

given data

Given data

depth = 2.60 m

density = 915 kg/m3

length = 5.00 m

width = 3.00 m

to find out

weight of the olive oil, force of air pressure and the pressure exerted upward

solution

we know density = mass / volume

mass = density* width *length *depth

mass = (915)(3)(5)(2.60)

mass = 3.57 x 10^4 Kg

so weight = mg = 3.57 x 10^4 (9.81) = 3.50 x 10^5 N

weight is 3.50 x 10^5 N

and

we know force = pressure * area

area = 3 *5 = 15 m²

and we know atmospheric Pressure is about 1.01 * 10^5 Pa

so force = 1.01 * 10^5 (15) = 1.52 x 10^6 N

force is 1.52 * 10^6 N

and

we know Fup - Fdown = Weight

so

Fup = 1.52 *  10^6 + 3.50 *  10^5

Fup = 1.87 * 10^6 N

so pressure = Fup / area

pressure =  1.87 * 10^6  / 15

pressure is 1.25 * 10^5 Pa


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Answers

Answer:

B.  Energy increases with decreasing wavelength and increasing frequency.

Explanation:

What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.

Answers

Answer:

option C

Explanation:

the correct answer is option C

When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.

According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.  

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In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))

Answers

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²

a)

The difference Δ g =  9.96 -9.72 =0.24  m/s²

The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n

b)

For first one :

Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage

For second  :

Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage

c)

The mean g(mean )

g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2

The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

Explanation:

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

Learn more about Percent Difference and Error here:

brainly.com/question/33836796

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A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?

Answers

Answer:

The initial momentum of the ball is 8 kg-m/s.

Explanation:

Given that,

Mass of the ball is 4 kg

Initial speed of the ball is 2 m/s

Force applied to the ball is 5 N for 3 seconds

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Answers

Answer:

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Explanation:

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A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

Answers

Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance =  100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

h_u= Object height = 2.1 cm

Lens Equation

(1)/(f)=(1)/(u)+(1)/(v)\n\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\n\Rightarrow (1)/(v)=(1)/(-23.9)-(1)/(100)\n\Rightarrow (1)/(v)=(-1239)/(23900) \n\Rightarrow v=(-23900)/(1239)=-19.29\ cm

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

m=-(v)/(u)\n\Rightarrow m=-(-19.29)/(100)\n\Rightarrow m=0.1929

m=(h_v)/(h_u)\n\Rightarrow 0.1929=(h_v)/(2.1)\n\Rightarrow h_v=0.1929* 2.1=0.40509\ cm

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.