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3 years ago

1-) Two blocks of mass 3.SO kg and 8.00 kg are connected by a massless string that passes over a

Physics

1 answer:

Troyanec [42]3 years ago

3 0

Answer:

The magnitude of the acceleration of each block is, a = 2.56 m/s²

The tension in the string is, T = 43.05 N

Explanation:

Given data,

The larger block of mass, M = 8.00 kg

The smaller block of mass, m = 3.50 kg

The formula for Atwood machine is,

Ma = Mg - T

ma = T - mg

Adding those equations,

a (M + m) = g ( M - m)

a = (M + m) / ( M - m)

Substituting the values,

a = (8 + 3.5) / (8 - 3.5)

= 2.56 m/s²

The magnitude of the acceleration of each block is, a = 2.56 m/s²

The tension in the string,

T = m(a + g)

= 3.5 ( 2.56 + 9.8)

= 43.05 N

The tension in the string is, T = 43.05 N

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Answer:

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2 years ago

A book falls of the table and hits the floor. Is work done? Why?

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3 years ago

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Suppose the average mass of each of 20,000 asteroids in the solar system is 1017 kg. Compare the total mass of these asteroids t

blondinia [14]

Answer:

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

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$\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184$

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

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$V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m$

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3 years ago

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3 years ago

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docker41 [41]

Answer:

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Explanation:

a) As this is a perfectly elastic collision, we can use the formula $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}$ , due $v_{2i}=0m/s$ , we obtain $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}$ . Then with the data that we know $m_{1}=5.21g=0.00521kg, m_{2}=14.8kg$ and $v_{1i}=412m/s$ , therefore $v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s$ or $v_{2f}=(0.289\±0.002)m/s$ adding uncertainty.

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4 years ago