B
Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is 
where a = 1.2 km north
and b = 1.6 km east
magnitude = 2 km
Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.
I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.
If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.
I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.
Just taking a wild guess here . . . . .
If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .
But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
Answer:
0.6kg
Explanation:
the unknown here is the mass of the second block
applying the law of the conservation of momentum
m₁v₁ + m₂v₂ = (m₁ + m₂) v₃
where m₁=mass of first block=2.2kg
m₂=mass of colliding block= ?
v₁= velocity of first block=1.2m/s
v₂=velocity of colliding block=4.0m/s
v₃= final velocity of combined block=1.8m/s
applying the formula above
(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8
2.64 + 4m₂ = 3.96 + 1.8m₂
collecting like terms
4m₂ - 1.8m₂ = 3.96 - 2.64
2.2m₂=1.32
divide both sides by 2.2
m₂= 0.6kg
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation =
sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
