How much force is required to accelerate a 100 kg wagon at 5 m/s2?

12. The red light emitted by a He-Ne laser has a wavelength of 6.33 x 10^-7 m in the air and travels 3.0 x 10^8 m/s. Find the fr

Ymorist [56]

Answer:

f = 4.76 × 10^14 Hz

Frequency of the laser light f = 4.76 × 10^14 Hz

Explanation:

Given;

wavelength = 6.33 x 10^-7 m

Speed of light v = 3.0 x 10^8 m/s.

Frequency of the laser light f

= speed of light/wavelength

Substituting the given values;

f = 3.0 x 10^8 m/s ÷ 6.33 x 10^-7 m

f = 4.76 × 10^14 Hz

Frequency of the laser light f = 4.76 × 10^14 Hz

4 0

4 years ago

A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then s

tino4ka555 [31]

Given:

k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs

Solution:

F = -kx
mx" = -kx
x" + (k/m)x = 0

characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)

x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)

ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)

x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0

1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft

x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0

So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[(2π/0.1107)t]

5 0

3 years ago

A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with

Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 + 20 sin (30) t - 4.9 t^2

when it hits ground this = 0

0 = -4.9t^2 + 20 sin (30 ) t + 45

0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec

max height is at t= - b/2a = 10/9.8 =1.02

use this value of 't' in the equation to calculate max height = 50.1 m

it has 4.22 - 1.02 to free fall = 3.2 seconds free fall

v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s

total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m

8 0

2 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: $t=\frac{0.5(2675.5-417.36)+(-0.0522)}{0.85}=1328.3(s)$

6 0

3 years ago

A gas is placed inside a very cold box. The gas molecules transfer heat to the walls of the box. Which best describes what happe

STatiana [176]

Answer:

the gas molecules slow down, and some change phase to liquid.

Explanation:

1) The heat always travels from the warmer substance to the coleer substace.

That is why it is correctly indicated that the gas molecules transfer heat to the walls of the box.

2) As this happens, means that the gas molecules lose heat energy and is cooled.

3) KInetic energy is directly related with the temperature: therefore, the lower the temperature the lower the kinetic energy.

4) Kinetic energy is, also, directly related to the square of the speed. So, the lower the kinetic energy, the lower the average speed of the molecules.

That is why it is correct to say that the gas molecules slow down.

5) Since it is said that the walls of the box are very cold, it may well happen that the energy passed from the molecules to the walls of the box is enough to produce the phase change, from gas to liquid.

As for the other options:

i) the gas molecules speed up due to collisions with the walls: No. As already stated the gas molecules slow down.

ii) the gas molecules vibrate around fixed positions: No, the gas molecules are not around fixed positions, they translate in right random motion.

iii) the gas molecules gain energy from the cold walls of the box: as already stated the gas molecules release energy which is transfered to the cold walls of the box.

8 0

3 years ago