Answer:

Explanation:
From the question we are told that:
Separation Distance 
Potential difference 
Generally the equation for Electric Field strength is mathematically given by



Answer:
It can occur only when light is incident on an interface where the index of refraction on the other side is less.
Explanation:
When the light passes from a denser medium, with refractive index n1, to another less dense medium, with refractive index n2, the incident light beam is refracted in such a way that it is not able to cross the surface between both media, the light beam is fully reflected and completely confining in the optically denser medium through which it propagates. For this phenomenon to occur, it is necessary that the angle of the incident light beam with respect to the normal be greater than or equal to the critical incidence angle θc. The critical angle can be calculated as :

Answer:
x = 4.32 [m]
Explanation:
We must divide this problem into three parts, in the first part we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force = 700 [N]
m = mass = 2030 [kg]
a = acceleration [m/s²]
Now replacing:
![F=m*a\\700=2030*a\\a = 0.344[m/s^{2}]](https://tex.z-dn.net/?f=F%3Dm%2Aa%5C%5C700%3D2030%2Aa%5C%5Ca%20%3D%200.344%5Bm%2Fs%5E%7B2%7D%5D)
Then we can determine the final speed using the principle of conservation of momentum and amount of movement.

where:
m₁ = mass of the car = 2030 [kg]
v₁ = velocity at the initial moment = 0 (the car starts from rest)
Imp₁₋₂ = The impulse or momentum (force by the time)
v₂ = final velocity after the impulse [m/s]
![(2030*0) + (700*5)=(2030*v_{2})\\3500 = 2030*v_{2}\\v_{2}=1.72[m/s]](https://tex.z-dn.net/?f=%282030%2A0%29%20%2B%20%28700%2A5%29%3D%282030%2Av_%7B2%7D%29%5C%5C3500%20%3D%202030%2Av_%7B2%7D%5C%5Cv_%7B2%7D%3D1.72%5Bm%2Fs%5D)
Now using the following equation of kinematics, we can determine the distance traveled.

where:
v₂ = final velocity = 1.72 [m/s]
v₁ = initial velocity = 0
a = acceleration = 0.344 [m/s²]
x = distance [m]
![1.72^{2}=0^{2} +(2*0.344*x) \\2.97 = 0.688*x\\x = 4.32 [m]](https://tex.z-dn.net/?f=1.72%5E%7B2%7D%3D0%5E%7B2%7D%20%2B%282%2A0.344%2Ax%29%20%5C%5C2.97%20%3D%200.688%2Ax%5C%5Cx%20%3D%204.32%20%5Bm%5D)
TheThe distance they have covered for trip is 219Km.
<h3>What do you mean by uniformly accelereted motion?</h3>
When an object is traveling in a straight line with an increase in velocity at equal intervals of time.
The total time for the trip is
T→ t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at
V1= 89.5 km/ h .
the distance traveled is ∆x = V1t1=Vavg T
after applying value and calculating it we get
t1= 2.44h for a total time of
t total we get 2.81 h .
∆x =V1T1 = VavgTtotal
∆x = 77×2.81= 219 Km
to learn more about Uniform accelereted motion click here brainly.com/question/12920060
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We use the binomial theorem to answer this question. Suppose we have a trinomial (a + b)ⁿ, we can determine any term to be:
[n!/(n-r)!r!] a^(r) b^(n-r)
a.) For x⁵y³, the variables are: x=a and y=b. We already know the exponents of the variables. So, we equate this with the form of the binomial theorem.
r = 5
n - r = 3
Solving for n,
n = 3 + 5 = 8
Therefore, the coefficient is equal to:
Coefficient = n!/(n-r)!r! = 8!/(8-5)!8! = 56
b.) For x³y⁵, the variables are: x=a and y=b. We already know the exponents of the variables. So, we equate this with the form of the binomial theorem.
r = 3
n - r = 5
Solving for n,
n = 5 + 3 = 8
Therefore, the coefficient is equal to:
Coefficient = n!/(n-r)!r! = 8!/(8-3)!8! = 56