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Julli [10]
2 years ago
10

A small sphere is hung by a string from the ceiling of a van. When the van is stationary, the sphere hangs vertically. However,

when the van accelerates, the sphere swings backward so that the string makes an angle of
θ
with respect to the vertical.
(a) Derive an expression for the magnitude a of the acceleration of the van in terms of the angle
θ
and the magnitude g of the acceleration due to gravity.
(b) Find the acceleration of the van when
θ=10.0∘
(c) What is the angle
θ
when the van moves with a constant velocity?
Physics
1 answer:
olganol [36]2 years ago
3 0

Answer with Explanation:

We are given that

String makes an angle w.r.t  vertical=\theta=

a.We have to derive an expression for the magnitude of the acceleration of the van in terms of the angle \theta and magnitude g of the acceleration due to gravity.

According to newton's second law

T sin\theta=ma

Tcos\theta=mg

\frac{Tsin\theta}{Tcos\theta}=\frac{ma}{mg}

tan\theta=\frac{a}{g}

a=gtan\theta

b.\theta=10^{\circ}

g=9.8 m/s^2

a=9.8\times tan10^{\circ}=1.73 m/s^2

c.Velocity=Constant

We have to find the angle \theta

a=0

0=9.8tan\theta

tan\theta=0

tan\theta=tan0

\theta=0^{\circ}

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2.4525 N

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A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but th
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15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow {V_2}=\frac{V_1}{T_1}\times T_2\\\Rightarrow {V_2}=\frac{12}{298.15}\times 373.15\\\Rightarrow {V_2}=15.01 L

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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
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Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

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