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swat32
3 years ago
12

A net force of 50 N causes a mass to accelerate at a rate of 6.8 m/s2. Determine the mass. ​

Physics
2 answers:
DochEvi [55]3 years ago
8 0

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ ‎ ‎‎‎ ‎\bigstar\boxed{\large\bf{\leadsto{7.35\:kg}}}

__________________________________________

\large\bf{\underline{Given:}}

  • Force = 50 N
  • Acceleration = \bf{6.8\:ms^{-2}}

\large\bf{\underline{According\:to\: formula:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\underline{\boxed{\large\sf\pink{F=M\times A}}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼50=M\times 6.8}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼M = \frac{50}{6.8}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼M= 7.35}

\large\mathfrak{Therefore,\:Mass=7.35\:kg}

Taya2010 [7]3 years ago
5 0
  • Force=50N
  • Acceleration=6.8m/s^2

\\ \sf\longmapsto F=ma

\\ \sf\longmapsto m=\dfrac{F}{a}

\\ \sf\longmapsto m=\dfrac{50}{6.8}

\\ \sf\longmapsto m=7.3kg

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Answer:

a.  9.99625 cm b. 68 °C

Explanation:

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Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

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We now find the length L₁ at T₁ = 0 °C from

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L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

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With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

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Answer:

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F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

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Recall that 1 J is equal to 6.242*10^{12}MeV, so:

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We can calculate v from the kinetic energy of the proton:

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