<span>Answer:
1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use.
Given: m= 1200kg v initial = 95km/hr v final = 0
2nd, focus on the units - in most cases units speak for the concept
the unit of the unknown is kcal, thus its the unit of energy or work
so, W = ?
3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present
since W = delta K.E =delta P.E
W= 0.5m( vf^2 - vi^2) ---> best formula
4th, Substitute the given to the formula
since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J
we express first 95 km/hr to m/s
95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec
W= 0.5(1200kg)[(0^2- (26.39m/sec)^2]
W=600 kg(0 - 696.43m^2/s^2)
W=600kg(-696.43m^2/s^2)
W=417859.3Nm or 417859.3 J
W = 417859.3 J x 1 cal /4.19 J
W = 99,727.7 cal or 99.728 kcal</span>
So first we find the gap between the slits by the formula d=1/N
<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>
<span>Now we use the formula dsin(angle)=n(wavelength) </span>
<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>
<span>so wavelenth=dsin(angle)/n </span>
<span>=[(2.7*10^-6)*sin30]/3 </span>
<span>=4.5*10^-7 m</span>
The period of the transverse wave from what we have here is 0.5
<h3>How to find the period of the transverse wave</h3>
The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.
The formula with which to get the period is
w = 4π
where w = 4 x 22/7
2π/T = 4π
6.2857/T = 12.57
From here we would have to cross multiply
6.2857 = 12.57T
divide through by 12.57
6.2857/12.57 = T
0.500 = T
Hence we can conclude that the value of T that can determine the period based on the question is 0.500.
Read more on transverse wave here
brainly.com/question/2516098
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If you have no idea what the voltage is that you're about to measure,
then you should set the meter to the highest range before you connect
it to the two points in the circuit.
Analog meters indicate the measurement by moving a physical needle
across a physical card with physical numbers printed on it. If the unknown
voltage happens to be 100 times the full range to which the meter is set,
then the needle may find itself trying to move to a position that's 100 times
past the highest number on the meter's face. You'll hear a soft 'twang',
followed by a louder 'CLICK'. Then you'll wonder why the meter has no
needle on it, and then you'll walk over to the other side of the room and
pick up the needle off the floor, and then you'll probably put the needle
in your pocket. That will end your voltage measurements for that day,
and certainly for that meter.
Been there.
Done that.
Explanation:
It is given that,
An electric dipole consists of charges +2e and -2e separated by 0.82 nm
Charge, 
Distance between charges, 
Electric field strength, 
(a) The magnitude of the torque on the dipole is given by :
When dipole moment is parallel to the electric field, 
(b) When the dipole is perpendicular to the electric field, 
(Since, p = q × d)
(c) When the dipole moment is anti parallel to the electric field, 
Since, 
Hence, this is the required solution.
