Answer:
230.4 s
Explanation:
The speed of car A is

and the distance travelled is

so the time taken for car A is

The speed of car B is

and the distance travelled is

so the time taken for car B is

So the difference in time is

Which corresponds to

so car B arrived 230.4 s before car A.
Answer:
b) 68,9 km/h a) picture
Explanation:
In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.
Using this formula we pass time to hours:

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.
Using the values of the speed we calculate the distance in each interval

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).
Finally, we compute the average speed with the distance over time:

Answer:
Δμ = hΔf/B
Explanation:
If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.
Since the magnetic energy change equals the photon energy,
ΔE = ΔU
hΔf = ΔμB
Δμ = hΔf/B
Answer:
(A) Reading will be 65 N
(B) Net force on the elevator will be 49.076 N
Explanation:
We have given the balance force = 65 N
Acceleration due to gravity 
We know that W=mg
So 
m = 6.632 kg
(a) In first case as the as the speed is constant so the force on the elevator will be 65 N
(B) In second case as the elevator is decelerating at a rate of 
So net acceleration = 9.8-2.4=
So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N
<h2><em>state coulombs law in word</em></h2>
- <em>: a statement in physics: <u>the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between </u></em><em><u>them</u></em>
<em><u>hope </u></em><em><u>it</u></em><em><u> helps</u></em>
<em><u>#</u></em><em><u>c</u></em><em><u>a</u></em><em><u>r</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>on</u></em><em><u> learning</u></em>