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nignag [31]
4 years ago
7

Each of 134 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

ock is pulled with a force of 134 N.
(a) What is the tension in the string connecting block 126 to block 127?
(b) What is the tension in the string connecting block 63 to block 64?
Physics
1 answer:
poizon [28]4 years ago
8 0

Answer:

A. T=126N

B. T=63N

Explanation:

To determine the tension in each given blocks, we first determine the acceleration of each block. It obvious that each mass will move with the same acceleration since the string connecting them is massless.

Hence using the equation of force we have

F=ma

Where m=total mass of blocks,

a=acceleration

F= force applied in this case the tension in the string.

For a 134 identical masses with an applied force of 134N, the acceleration of each mass can be computed as

134=134m*a

a=134/134m

a=(1/m )m/s²

a. To calculate the tension in the string between the 126 and 127 block, we use the equation below

T=ma

Since the number of blocks before the string is 126, we multiply the mass of each block by 126.

Hence the tension can be computed as

T=126m*a

Since a=1/m then

T=126m*1/m

T=126N

B.To calculate the tension in the string between the 63 and 64 block, we use the equation below

T=ma

Since the number of blocks before the string is 63, we multiply the mass of each block by 63.

Hence the tension can be computed as

T=63m*a

Since a=1/m then

T=63m*1/m

T=63N

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Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
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Answer

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\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

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      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

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      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

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