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denis-greek [22]
3 years ago
10

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

op half is filled with oil of specific gravity of 0.85. What is the pressure difference between the top and bottom of the container?
Physics
2 answers:
Step2247 [10]3 years ago
8 0

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

tamaranim1 [39]3 years ago
5 0

Answer:

I don't get it I don't get it

Explanation:

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Pepsi [2]

Answer:

The mini Cooper will experience the greater force

Explanation:

Generally, a bulldozer has a greater mass compared to a Mini Cooper hence when both of these vehicles interact in an head on collision the Mini Cooper will experience a greater force because the bulldozer has a greater momentum

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a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum
7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 ×10^{-2} m

height (h) =8.8 cm = 8.8 ×10^{-2} m

total charge (q) =6.1×10^{-7} C

To solve:

volume charge density (ρ)

We know;

<u> ρ =q ÷ v</u>

volume of cylinder = π ×r  × r × h

volume of cylinder =3.14 × 2.1 × 2.1 ×10^{-4} × 8.8 ×10^{-2}

volume of cylinder (v) = 122.23 ×10^{-6}

<u> ρ =q ÷ v</u>

ρ = 6.1×10^{-7} ÷   122.23 ×10^{-6}

<u>ρ = 0.04 × </u>10^{-1}<u> C</u>

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

3 0
3 years ago
Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.
andriy [413]
The answer I believe is False
8 0
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The downward force produced when air flows over the winglike spoiler on a race car is an example of ___ principle
Murljashka [212]

As per bernoulli's principle

P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2

here

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P_2 = pressure downwards

v_1 = velocity of air upwards

v_2 = velocity of air downwards

now from this equation we can say that the pressure difference will be

P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2

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5 0
3 years ago
Read 2 more answers
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
2 years ago
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