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Ksivusya [100]
3 years ago
12

a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

is 16.0 cm high the oil air interfdace is 4.5 cm above the liquid level in the left arm of the u tube algrebraic expression to deter,ine the density of the unknown fluid
Physics
1 answer:
scoray [572]3 years ago
8 0

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
ra1l [238]

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   <em>(I)</em>

Car

x = x0 + v0*t

x = 52 + 21.5t  <em>(II)</em>

<em />

then we can apply <em>I = II</em>

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use <em>I</em> or <em>II</em>, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

8 0
3 years ago
A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m
Whitepunk [10]

Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

3 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
oee [108]

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

3 0
3 years ago
A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl
nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0
3 years ago
An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel
Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

                  S = ut + ½ gt²

∴                 g = 2S/t²

Substituting the given values,

                   g = 2 x 144 /6²

                      = 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0
3 years ago
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