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What is the length of a pendulum with a period of one second?

worty [1.4K]

Answer:

A simple pendulum with a period of 1 second will have a length of 0.25 meters or 25 centimeters

Explanation:

hope it helps u☺☺

7 0

3 years ago

II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizont

Lostsunrise [7]

Hi there!

In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.

The net force is equivalent to the centripetal force, so:

∑F = T

mv²/r = T

Solve for v:

v = √rT/m

v = 13.96 m/s

3 0

3 years ago

An open cooking pot containing 0.5 liter of water at 20°C, 1 bar sits on a stove burner. Once the burner is turned on, the water

sesenic [268]

Answer:

a. the time required for the onset of evaporation is: 196.1 seconds and b. the time required for all of the water to evaporate is: 1328.3 seconds.

Explanation:

We need to stablish that there is 3 states at this problem. At the firts one, water is compressed liquid and the conditions for this state are: P1=100KPa,T1=20°C,V1=0.5m^3. From the compressed liquid chart and using extrapolation, we can get: v1=vf1=0.0010018 (m^3/Kg) and u1=uf1=83.95(KJ/kg). Now we can find the mass of water at the state 1 as: $m=\frac{V_{1} }{v_{1} } =\frac{0.5*10^{-3} }{0.0010018}=0.5(Kg)$ Then the liquid water is heated at a rate of 0.85KW, and its volume increase, while work is done by the system at the boundary, we can assume that the pressure remains constant throughout the entire process. At the second state the water is saturated liquid and the conditions are: P2=100KPa, T2=Tsat=99.63°C, v2=vf2=0.001043(m^3/Kg) and u2=uf2=417.36(KJ/Kg). Now we can find the work as: $W=mP(v_{2} -v_{1} )=0.5*100*(0.001043-0.0010018)=0.00207(KJ)$ . (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or $m(u_{2} -u_{1} )= Q*t-W$ , solving for t we get the time required for the onset of evaporation: $t=\frac{0.5*(417.36-83.95)+0.00207}{0.85}=196.1(s)$ .(b) Then continue heat transfer to the cooking pot and results in phase change getting vapor at 99.63°C. At the final state or third state the mass is zero because all liquid was evaporated and the initial mass at this state is the same for the second state: 0.5 (Kg) and doing an energy balances results in: $(m_{3} u_{3} -m_{2} u_{2})=Q*t-W+( m_{3}-m_{2})h_{e}$ , but m3=0, now solving for t we can get the time required for all of the water to evaporate as: $t=\frac{m_{2}(h_{e}-u_{2})+W}{Q}$ . We can get from the saturated liquid chart the enthalpy he=hge=2675.5(KJ/Kg) @P=100KPa. Now we need to calculate the work related with the volume decreases as vapor exits the control volume or process 2-3 work boundary as: $W=\int\limits^3_2 {p} \, dV= p*(V_{3} -V_{2} )=-m_{2} P_{2} v_{2} =-(0.5)*100*0.001043=-0.0522(KJ)$ . Now replacing every value in the time equation we get: $t=\frac{0.5(2675.5-417.36)+(-0.0522)}{0.85}=1328.3(s)$