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3 years ago

Heat in a substance is an example of _____ energy?

1 answer:

3 0

Heat in a substance is still an example of heat or thermal energy. Heat or thermal energy is further defined as the amount of available molecules of kinetic energy transferred in a specific amount of time. Heat is nothing if it is not transferred from one object to another. There are many uses of heat energy. When we are cooking using our stoves, there is heat energy. When we pour hot water onto our cup of coffee, there is heat energy.

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For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.96 x 10³ eV

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 4.19 x 10⁴ eV

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 3.73 x 10⁹ eV

8 0

2 years ago

2.4 physics assignment.

kvv77 [185]

Answer:The distance o the ramp that the car traveled is given by d=(1/2)at^2=(0.5)(3.96)(5.76)^2=65.69 meters. The horizontal component of this travel is 65.69*

Explanation:

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2 years ago

The device used to measure a masses of a body is kilogram . true or false

olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

6 0

3 years ago

16) A man ran a 5 mile race. The race looped around a city park and back

finlep [7]

Answer:

The man's total displacement is equal to 0.

Explanation:

Given that,

A man ran a 5 mile race. The race looped around a city park and back to the starting line.

We need to find the total displacement of the man.

We know that,

Displacement = shortest path covered

Also,

Displacement = final position - initial position

As it reaches back to its starting line, it means, the displacement is equal to 0.

Hence, the man's total displacement is equal to 0.

3 0

3 years ago

When you flip a penny (2.35 g), it leaves your hand and moves upward at 2.85 m/s. Use energy to find how high the penny goes abo

Kaylis [27]

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

8 0

3 years ago