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just olya [345]
3 years ago
6

A 15.0 m long steel rod expands when its temperature rises from 34.0 degrees C to 50.0 degrees C. What is the change in the beam

's length due to the thermal expansion? Steel has a coefficient of linear expansion of 12.0 E -6/C degrees.
0.0612 m
0.00432 m
0.00288 m
0.0119 m
0.0475 m
Physics
2 answers:
azamat3 years ago
8 0
0.00288................
Lilit [14]3 years ago
5 0

Answer:

The change in beam's length due to the thermal expansion is 0.00288 meters.

Explanation:

Given that,

Original length of the steel rod, l = 15 m

Initial temperature, T_i=34^{\circ} C

Final temperature, T_f=50^{\circ} C

The coefficient of linear expansion of the steel, \alpha =12\times 10^{-6}\ /^{\circ} C

Let \Delta l is the change in beam's length due to the thermal expansion. It can be given by :

\dfrac{\Delta l}{l}=\alpha (T_f-T_i)\\\\\Delta l=l\alpha (T_f-T_i)\\\\\Delta l=15\times 12\times 10^{-6}\times (50-34)\\\\\Delta l=0.00288\ m

So, the change in beam's length due to the thermal expansion is 0.00288 meters. Hence, this is the required solution.

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The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.  

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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t
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Answer:

Explanation:

Given

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r_A=-80sin35\hat{i}+80cos35\hat{j}

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Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

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v_B=20sin80\hat{i}+20cos80\hat{j}

Velocity of A w.r.t B

v_{AB}=v_A-v_B

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4 0
3 years ago
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2 years ago
A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
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We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

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From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

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4 0
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An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov
jeka94

Answer:

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Explanation:

As we know that there is no friction on the system or there is no external force on this system

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now we have

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4 0
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