Ask question

Ask question

All categories

3 years ago

12

How high would the level be in a gasoline barometer at normal atmospheric pressure?

1 answer:

Sergio [31]3 years ago

7 0

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

P = S.G*p_w*g*h

Where, h = P / S.G*p_w*g

- Input the values given:

h = 101.325 KPa / 0.739*9.81*997

h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

You might be interested in

Question 3 of 10

Stels [109]

Answer:

30N

Explanation:

6 0

3 years ago

_________ refers to conversion in the context of intentional torts against property?

Sveta_85 [38]

Answer:

A tort that permanently interferes with another's use and enjoyment of his or her personal property.

Explanation:

An intentional tort is a category of wrongful acts done on purpose by the tortfeasor. Examples are trespass to chattels, conversion, false imprisonment, battery, defamation and assault.

A tort that permanently interferes with another's use and enjoyment of his or her personal property refers to conversion in the context of intentional torts against property.

5 0

3 years ago

Read 2 more answers

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

Question 8 a-e plz

N76 [4]

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

5 0

3 years ago

All objects emit electromagnetic radiation as a result of _____.

Artist 52 [7]

One

All objects, every one, as long as the object is not at absolute zero (not very many objects are -- like no common object) emit electromagnetic radiation because they possess thermal (heat) energy.

The answer, surprisingly, is A.

Two

This is just an example of Snells law. The angles are with the normal.

Index of Refraction = sin(input angle) / sin(outgoing angle)

Index of Refraction = sin(angle incidence)/sin(angle of refraction)

Index of Refraction = Sin (36)/Sin(27.5)

Index = 1.28 (rounded)

8 0

3 years ago