Answer:
v = 0.4712 m / s In the opposite direction of the turntable
Explanation:
For this exercise the mouse to remain motionless must run at the same speed as the record player point moves,
Let's use rotational kinematics to find the speed of the record player at the point where the mouse is
v = w r
Let's reduce the magnitude to the SI system
w = 45 rpm (2π rad / 1rev) (1 min / 60s) = 4.712 rad / s
r = 10 cm = 0.10 m
Let's calculate
v = 4.712 0.10
v = 0.4712 m / s
In the opposite direction of the turntable
Answer:
i would like to help you but I can't help you i don't understand the question and i don't speak Spanish
Answer:
(a) 0.81 V
(b) 0.52 V
Explanation:
Number of turns, N = 150
Radius, r = 2.5 cm = 0.025 m
Magnetic field, B = 0.060 T
f = 440 rev/min = 440 / 60 = 7.33 rps
A.
The maximum emf is given by
e = N x B x A x 2 x π x f
e = 150 x 0.060 x 3.14 x 0.025 x 0.025 x 2 x 3.14 x 7.33
e = 0.81 V
B.
The back emf is given by
e' = 2e / π = 2 x 0.81 / 3.14 = 0.52 V
<span>The correct answer is option B. i.e UV Rays. The Ultra violet light is the part of electromagnetic light whose wavelength is less than the Visible light light. i.e. UV range is from 10 nm to 400 nm. UV light is invisible to the naked eye.</span>
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: 
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:
. (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or 
, solving for t we get the time required for the onset of evaporation:
.(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:
, but m3=0, now solving for t we can get the time required for all of the water to evaporate as:
. 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: 
. Now replacing every value in the time equation we get:
