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mario62 [17]
3 years ago
12

For each of the situations described below, the object considered is undergoing some changes. Among the possible changes you sho

uld consider are:
(Q) The object is absorbing or giving off heat.
(T) The object's temperature is changing.
(E) The object's thermal energy is changing.
(W) The object is doing mechanical work or having work done on it.
(C) The objects chemical energy is changing.

For each of the two situations, identify which of the five changes is taking place, indicating the applicable letters Q T U W C (or none) as appropriate while providing your reasoning.

a. An ice cube sits in the open air and is melting.
b. A cylinder with a piston on top contains a compressed gas and is sitting on a thermal reservoir (a large iron block that can provide or take thermal energy as the system needs). After everything has come to thermal equilibrium, the piston is moved upward somewhat (very slowly). The object to be considered is the gas in the cylinder.
Physics
1 answer:
velikii [3]3 years ago
8 0

Answer: a. (Q), (T), (E)

              b. (Q), (T), (E), (W)

Explanation: <u>Thermal</u> <u>Energy</u> and Temperature are closely related: when the temperature rises causing atoms or molecules to move, thermal energy is produced. Thermal energy is the energy within the system.

<u>Mechanical</u> <u>Work</u> is the amount of energy transferred due to an applied force.

<u>Chemical</u> <u>Energy</u> is the energy contained in the bonds of chemical structures of the molecules released when a chemical reaction happens.

Given the explanations, let's analyse the situations:

a. For an ice cube to be melting, it has to be absorbing heat, which means its thermal energy is changing and, consequently, so does its temperature;

b. First, the gas inside the cylinder reaches a thermal equilibrium, which means its thermal energy and temperature changed. Since there were  exchange of heat to reach the equilibrium, the gas absorbed or gave off heat. After the equilibrium, when the piston starts to be moved, the energy of the pressure is transferred to the gas, so mechanical work had been done.

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0.22 L of pancake syrup has a mass of 33 g.
katrin2010 [14]

Answer:

a. 150 g/L

b. 75 g

c. 120 mL

Explanation:

a. 33g/0.22L=150 g/L

b. 33g/0.22L=150 g/L

150 g/L*0.5L=75g

c. 0.22L/33g=0.006667L/g

0.006667L/g*18g=0.12L

0.12L*1000=120mL

6 0
3 years ago
2. How much do you think the force of friction must be? Why?
Tamiku [17]

Answer:

It must be high do to the gravity

Explanation:

4 0
2 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b
givi [52]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The equivalent gravitational force is ~

  • F  \approx1.48\times 10 {}^{ - 7}  \: \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where,

  • F = gravitational force

  • m_1 = mass of 1st object = 500 kg

  • m_2 = mass of 2nd object = 20kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 2.12 m

Let's calculate the force ~

  • F = \dfrac{6.674   \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }

  • F = \dfrac{6.674  \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}

  • F =  \dfrac{6.674}{4.4944}  \times 10 {}^{ - 7}

  • F =1.484 \times 10 {}^{ - 7}  \: \: newtons
7 0
2 years ago
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the
lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0
3 years ago
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