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qwelly [4]
3 years ago
5

On a smooth horizontal floor, an object slides into a spring which is attached to another mass that is initially stationary. Whe

n the spring is most compressed, both objects are moving at the same speed. Ignoring friction, and considering the two masses and spring as a system, what is conserved during this interaction?A) momentum and potential energyB) kinetic energy onlyC) momentum and kinetic energyD) momentum onlyE) momentum and mechanical energy
Physics
1 answer:
HACTEHA [7]3 years ago
3 0

Answer:

E) momentum and mechanical energy

Explanation:

In the context, an object is attached to the another mass with a spring which is initially at a rest position. Now when the spring is compressed, the two masses moves with the same speed. Now since the both the masses combines with the spring to move together they are considered as one system and in this case the momentum and the kinetic energy will be conserved.

The kinetic energy and momentum of the system after collision and the kinetic energy and momentum of the two masses before collision will be constant.

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From part a, you know that surface temperature is a stellar property that we infer indirectly. What must we measure directly so
trasher [3.6K]

We need to directly measure the spectral type in order to determine the surface temperature of a star.

<h3>How do you find the properties of a star?</h3>

Astronomers can determine the temperature of a star by looking at its color and spectrum. The apparent brightness of a star describes how luminous it looks to us. The brightness of a star tells us how bright it really is. The luminance can be determined using both the perceived brightness and the distance.

A star's luminosity, or the total amount of energy it emits each second, is determined by two factors: The stellar photosphere's "Effective Temperature," T. the star's total surface area, which is influenced by its radius, R.

Because it controls how much fuel a star has and how quickly it burns it, a star's mass is its most fundamental characteristic. The majority of a star's life is spent burning hydrogen into helium in its core, which generates energy. The star needs to achieve a balance between gravity and outward pressure in order to continue to be "alive."

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4 0
1 year ago
Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for
klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

  • A is amplitude of the wave
  • C is phase difference of the wave

<h3>What is angular frequency of a wave?</h3>

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

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5 0
1 year ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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3 years ago
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Answer b protons and electrons
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Vladimir79 [104]

Answer:

Gases, liquids and solids are all made up of microscopic particles, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

<h3>Hope this is fine for you</h3>
6 0
3 years ago
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