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vodomira [7]
3 years ago
15

What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?

Physics
1 answer:
goblinko [34]3 years ago
6 0
In physics, Hooke's law is written in equation as:

F = kx

It states that the force F exerted on the spring is directly proportional to the displacement x by a constant called spring constant k.

In the laboratory, this is done in an experiment through the apparatus shown in the attached figure. The object experimented here is the spring, and you are to find the spring constant. A known mass of object is attached below the spring. That object carries a force in the form of gravitational pull in terms of weight. When the spring stretches, the displacement is measured with the use of the ruler.

There are a number of sources of error for this experiment. First, the reading from the ruler by the reader may be inaccurate. That's why digital balances are much more reliable because it minimizes human error. Reading the measurement on the ruler is subjective especially when you don't read it on eye level. Second, the force of the object might also be inaccurate if you use an unreliable weighing scale. Lastly, the apparatus might not be properly calibrated.

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A person is listening with his ear against the rail for an oncoming train. When the train is 1.65 km away, how long will it take
Vitek1552 [10]

Answer:

Time = 0.317 seconds.

Explanation:

Given the following data;

Distance = 1.65km to meters = 1.65 * 1000 = 1650 meters

We know that the speed of sound in steel is equal to 5200m/s

To find the time to hear the sound of the whistle;

Time = distance/speed

Substituting into the equation, we have

Time = 1650/5200

Time = 0.317 seconds.

Therefore, it will take him 0.317 seconds to hear the sound of the whistle.

3 0
3 years ago
(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less
Free_Kalibri [48]

Answer:

q = 7.4 10⁻¹⁰ C

Explanation:

a) The magnetic force is given by the expression

        F = q v x B

Where the blacks indicate vectors, q is the electric charge, v at particle velocity and B the magnitude of the magnetic field. If the velocity is perpendicular to the magnetic field, the sine is 1

      F = q v B

Let's calculate the charge

      q = F / vB

      q = 1.00 10⁻¹² / 30.0 B

For the magnetic field of the earth we have a value between 25μT and 65μT, an intermediate value would be 45 μT, let's use this value.

     q = 1 10⁻¹² / (30 45 10⁻⁶)

    q = 7.4 10⁻¹⁰ C

b) In laboratories and modern electronics, currents of up to 1 10⁻⁶ A can be achieved without much difficulty, in advanced and research laboratories currents of up to 1 10⁻¹² can be managed. Load values ​​(coulomb) cannot they are widely used today for work, but 1 mA = 3.6C, so we see that getting loads with the value of 10⁻¹⁰ C implies very small current less than 1 10⁻¹³ A, which only in laboratories of Very specialized can be created. Consequently, from the above it would be difficult to find loads lower than the calculated

The electrostatic charge is the one created by the friction between two surfaces, it is an indicated charge, in this case it would be possible to have better wing loads found from 10⁻¹⁰C

4 0
3 years ago
Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of
marusya05 [52]

Answer:

a) \dot m = 6.878\,\frac{kg}{s}, b) T = 104.3^{\textdegree}C, c) \dot S_{gen} = 11.8\,\frac{kW}{K}

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The mass flow rate is:

\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

P = 7000\,kPa

T = 500^{\textdegree}C

h = 3411.4\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

State 2s - Liquid-Vapor Mixture

P = 100\,kPa

h = 2467.32\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

x = 0.908

The isentropic efficiency is given by the following expression:

\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}

The real specific enthalpy at outlet is:

h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})

h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )

h_{2} = 2684.46\,\frac{kJ}{kg}

State 2 - Superheated Vapor

P = 100\,kPa

T = 104.3^{\textdegree}C

h = 2684.46\,\frac{kJ}{kg}

s = 7.3829\,\frac{kJ}{kg\cdot K}

The mass flow rate is:

\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}

\dot m = 6.878\,\frac{kg}{s}

b) The temperature at the turbine exit is:

T = 104.3^{\textdegree}C

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0

\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})

\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )

\dot S_{gen} = 11.8\,\frac{kW}{K}

4 0
4 years ago
PLZZZ I will give brainliest. A ball that contains mechanical energy is rolled across the floor. You notice the ball is slowing
Oksanka [162]

Answer: because of friction it will stop rolling completly

Explanation:

8 0
3 years ago
What is meant by significant figures of a measurement ?
juin [17]

Answer:

The number of digits used to express a measured or calculated quantity. By using significant figures, we can show how precise a number is

Explanation:

6 0
3 years ago
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