Which energy store is increased when an object is heated?

Alona [7]

Yes indeed ! Very good !

3 0

3 years ago

The acceleration due to gravity on or near the surface of Earth is 32 ft./s/s. Neglecting friction, from what height must a ston

svetoff [14.1K]

Given :

The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s

To Find :

From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.

Solution :

Initial velocity of stone, u = 0 ft/s.

Now, by equation of motion :

$2as = v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft$

Therefore, height from which stone is thrown is 289 ft.

3 0

3 years ago

g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b

Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c) Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

Em₀ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₀ = Em_f

Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

Em₁ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₁ = Em_f

Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0

3 years ago

Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.1

ludmilkaskok [199]

Answer:

$1.189eV$

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

$U=\frac{kq_{e}q_{p}}{r}$

where $q_{e}$ is the electron charge, $q_{p}$ is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

$\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV$

5 0

4 years ago

When two plates collide, what determines which plate comes out on top

inn [45]

When two plates collide, it is the density that determines which plate comes out on top. The more denser plate subducts (goes under) the less denser plate. Hope this answers the question. Have a nice day. Feel free to ask more questions.

7 0

3 years ago