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Kazeer [188]
3 years ago
8

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

inear expansion coefficient of aluminum is 23.0 x 10^-6 /C°.
Physics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient \alpha=23.0\times10^{-6}/C^{\circ}

We need to calculate the volume expansion coefficient

Using formula of  volume expansion coefficient

\beta=3\alpha

Put the value into the formula

\beta=3\times23.0\times10^{-6}

\beta=0.000069=69\times10^{-6}/C^{\circ}

We need to calculate the volume

V= a^3

V=4^3

V=64\ cm^3

The change temperature of the cube is

\Delta T=T_{f}-T_{i}

Put the value into the formula

\Delta T=67-19 = 48^{\circ}C

We need to calculate the increases volume

Using formula of increases volume

\Delta V=V\beta\Delta T

Put the value into the formula

\Delta V=64\times69\times10^{-6}\times48

\Delta V=0.212\ cm^3

Hence, The volume of an aluminum cube is 0.212 cm³.

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Answer:

a) 633.39 J

b) 0.28

Explanation:

a)The kinetic energy of the player = \frac{1}{2} mv^{2}

Work done by friction = energy change of the player

                                    = \frac{1}{2} 67(4.35)^{2} = 633.9 J

b) Assuming the frictional force stays constant,

Work done by friction = Frictional force×distance

Frictional force = kinetic friction(μ)×normal reaction

Normal reaction = weight = mass×gravitational acceleration ( g=10m/s2 )

Combining these equations

633.9  =  F×3.4 ⇒ F = 186.44 N

F = μmg ⇒ μ = F/mg

                     = 186.44/670

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3 years ago
Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

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A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f
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Answer:

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\theta = Angle

t = Time taken

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\epsilon=NA\dfrac{d\phi}{dt}\\\Rightarrow \epsilon=\dfrac{NBA}{\Delta t}

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