Answer:
Explanation:
The formula to find average acceleration is a=Δv/Δt.
Convert 216 km/hr to m/s
216 km/hr * 1000 m/km * 1 hr/60 mins * 1 min/60 secs = 60 m/s
So a= (60 m/s)/(5.0*10^-3) = 12000 m/s^2
Put that into scientific notation to get 1.2*10^4 m/s^2
Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
Answer:
Change in electric potential energy is -28.0 J
Explanation:
Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.
OR
Electric potential energy is also equal to the change in the configuration of the charge particles.
Thus,
Change in electric potential energy = - Work Done
According to the problem, Work Done is equal to 28 J. Thus,
Change in electric potential energy = -28 J
Answer:
the no. of ejected electrons per second will increase.
Explanation:
In photoelectric effect, when a light is incident on a metal surface it ejects some electrons from the metal surface. The energy of photon of light must be equal to or greater than the work function of that metal. All the extra energy above the work potential appears as the kinetic energy of the ejected electrons. So, greater he energy of photon greater will be the kinetic energy of the ejected electrons.
A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, <u>the no. of ejected electrons per second will increase.</u>