B) millimeter, centimeter, meter, kilometer
Answer:

34.46 V/m
Explanation:
= Vacuum permeability = 
c = Speed of light = 
I = Intensity = 1.575 W/m²
The maximum magnetic field intensity is given by

The magnetic field intensity is 
The maximum electric field intensity is given by

The electric field intensity is 34.46 V/m
Permanent magnet. An induced magnet would be created when a piece of iron (for example) is in contact with a magnet. Temporary magnets would be something like an electromagnet. Bar magnets are permanently magnetic unless we heat them or hammer them to cause their domains to loose alignment.
Answer:
(a) 61.25 N
(b) 6.25 kg
(c) 6.25 Kg
Explanation:
Weight on moon = 10 N
Acceleration due to gravity on moon = 1.6 m/s^2
Acceleration due to gravity on earth = 9.8 m/s^2
Let m be the mass of the package.
(a) Weight on earth = mass x acceleration due to gravity on earth
Weight on earth = 6.25 x 9.8 = 61.25 N
(b) Weight on moon = mass x acceleration due to gravity on moon
10 = m x 1.6
m = 6.25 kg
(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct