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jek_recluse [69]
3 years ago
6

A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o

f copper that is initially at 85°C, and the other is a chunk of aluminum that is initially at 5.0°C. To the surprise of the student, the water reaches a final temperature of 25°C, precisely where it started. What is the mass of the aluminum chunk?
Physics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container =  120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, M_{cu} = 200 g

initial temperature of the copper cube, T_c_u = 85°C

initial temperature of the aluminum chunk T_A_l = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}

where;

C_{AL} is specific heat capacity of aluminum

\delta T_{Al} is change in temperature of aluminum

C_c_u is the specific heat capacity of copper

\delta T_c_u is the change in temperature of copper

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg

Therefore, the mass of the aluminum chunk is 258 g

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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0
4 years ago
Would you expect the bonds in ammonia to be polar covalent?
azamat
Yes I would expect them too
8 0
3 years ago
What is needed to set a body in motion
kumpel [21]

Answer:

Newtons law

Explanation:

According to this law, a body at rest tends to stay at rest, and a body in motion tends to stay in motion, unless acted on by a net external force.

3 0
3 years ago
What's going to happen in 3 weeks ?
Nana76 [90]

<u><em>Answer:</em></u>

<u><em>god knows.</em></u>

Explanation:

7 0
3 years ago
Read 2 more answers
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
3 years ago
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