Behavior has at least six dimensions, which are: frequency, duration, latency, topography, locus, and force. Since the coach is recording how long it takes, the track coach is recording the duration behavior because duration is a synonym for time. Duration is your answer.
Answer:
See the answers below.
Explanation:
We will solve this problem by calculating each part separately.
A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days.
Energy can be calculated by multiplying the value of the power of the equipment by the amount of time of use.
![500 [W]*[\frac{6min}{1day} ]*[\frac{1day}{24hr} ]*[\frac{1hr}{60min} ]=2.083 [W]](https://tex.z-dn.net/?f=500%20%5BW%5D%2A%5B%5Cfrac%7B6min%7D%7B1day%7D%20%5D%2A%5B%5Cfrac%7B1day%7D%7B24hr%7D%20%5D%2A%5B%5Cfrac%7B1hr%7D%7B60min%7D%20%5D%3D2.083%20%5BW%5D)
The cots of electricity is 5.6 cents per kWh. How much would it cost to operate the laptop for 24 hours a day for one week?
We know that the power of the latop is 75 [W], then we can calculate the cost, multiplying the value of the power by the value of the cost by the time of use of the computer.
![0.075[kW]*5.6[\frac{cents}{kw*h}}]*[\frac{24hr}{1day}]*[1week]*[\frac{7days}{1week} ]=70.56 [cents]](https://tex.z-dn.net/?f=0.075%5BkW%5D%2A5.6%5B%5Cfrac%7Bcents%7D%7Bkw%2Ah%7D%7D%5D%2A%5B%5Cfrac%7B24hr%7D%7B1day%7D%5D%2A%5B1week%5D%2A%5B%5Cfrac%7B7days%7D%7B1week%7D%20%5D%3D70.56%20%5Bcents%5D)
A toaster oven is 85% efficient. It uses 1200 J of energy. How much thermal energy is it producing?
Efficiency is defined as the relationship between the energy obtained on the energy delivered. Almost always the energy delivered is greater than the energy obtained (first law of thermodynamics).
Therefore.
![Effic = E_{obtained}/E_{delivered}\\0.85=E_{obtained}/1200\\E_{obtained}=1020[J]](https://tex.z-dn.net/?f=Effic%20%3D%20E_%7Bobtained%7D%2FE_%7Bdelivered%7D%5C%5C0.85%3DE_%7Bobtained%7D%2F1200%5C%5CE_%7Bobtained%7D%3D1020%5BJ%5D)
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
I’m assuming you’re supposed to calculate the resultant force?
425N (right) -300N (left)
=125 N to the right
Net force acting on mass = 20 - 15 = 5N. ( subtracted cuz friction always opposes the motion i.e it always acts in direction opposite to the motion of the object). According to Newton's 2nd law of motion, F(net) = ma. a =F (net) / m = 5/10 = 0.5 m/s^2. Hope it helps :)