Answer:
56 J
Explanation:
The following data were obtained from the question:
Energy 1 (E₁) = 7 J
Extention 1 (e₁) = 1.8 cm
Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm
Energy 2 (E₂) =?
Energy stored in a spring is given by the following equation:
E = ke²
Where E is the energy.
K is the spring constant.
e is the extension.
E = ke²
Divide both side by e²
K = E/e²
Thus,
E₁/e₁² = E₂/e₂²
7/ 1.8² = E₂/ 5.4²
7 / 3.24 = E₂/ 29.16
Cross multiply
3.24 × E₂ = 7 × 29.16
3.24 × E₂ = 204.12
Divide both side by 3.24
E₂ = 204.12 / 3.24
E₂ = 63 J
Thus, the additional energy required can be obtained as follow:
Energy 1 (E₁) = 7 J
Energy 2 (E₂) = 63 J
Additional energy = 63 – 7
Additional energy = 56 J
Because they come came out of the same thing so duh
Do the data support or refute the hypothesis, normally, the data for the first part of the experiment support the first hypothesis.
<h3>What is a hypothesis test?</h3>
Hypothesis testing refers to the formal procedures used by statisticians to accept or reject these hypotheses. When trying to make decisions, it is convenient to formulate assumptions or conjectures about the populations of interest, which consist of considerations about their parameters.
As a result of this, we can see that from the complete text, there is a set of data which is used to show the force that is applied to a cart that causes the acceleration of the cart to increase which supports Newton's second law.
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They will feel equally hot or cold at the same temperature.
What do you mean by temperature?
Temperature is a physical parameter that describes how hot stuff or radiation is. There are 3 different types of temperature scales: those described in regards to the average translational kinetic energy per freely moving microscopic substance in a body, such as a atom, compound, or anion, such as the SI scale; those that rely solely on purely <u>macroscopic properties and thermodynamic principles</u>, such as Kelvin's original definition; <u>and those that are not defined by theoretical principles, but are defined by convenient empirical properties of particulate matter.</u>
If they are the same temperature as the area of your body that is touching them, they will feel equally hot or cold. There will be no heat movement if the wood and iron at the same temperature as the skin, and the iron's higher thermal conductivity will be irrelevant. <u>The </u><u>specific temperature</u><u> cannot be specified because if you touch the materials with your fingertips, the wood and iron must match that</u><u> temperature</u><u>, and if you test them with your tongue, which is often considerably </u><u>warmer </u><u>than your hands, the wood and iron must be </u><u>warmer</u><u>.</u>
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