Two points, point A and point B, are situated above a current-carrying wire. Point B is located at a distance R from the wire, w
1 answer:
Explanation:
It is given that, there are two points situated above a current-carrying wire. Point B is located at a distance R from the wire, which is twice as far from the wire as point A such that,
And
We know that the magnetic field at a distance R is inversely proportional to the distance from wire as :
So, the magnetic field at point A larger than the magnetic field at point B by a factor of 2. Hence, this is the required solution.
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The dimensions of the rectangle are:
l = 50 m
b = m
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.
<h3>How do we find a perimeter of a rectangle?</h3>The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.
<h3>Given:</h3>As per the question:
Perimeter of the room is given as P = 200 m
The region is rectangular having a semicircle at each end.
Now,
Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.
Then, Area of the given rectangle, A = lb
Perimeter of the room, P is =
Therefore,
Now,
Area, A =
Now, differentiate A w.r.t l:
Again differentiating w.r.t 'l', we get:
< 0
Thus we get maximum are when
Therefore,
l = 50 m
Now, from
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The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
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