Answer:
a. one line down one line to the right one live to the northwest from the object
b. t1=190 t2=310
Explanation:
Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

Potential energy shifts:


Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.



This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
Answer:
1. W = F d = 20 N * 6 m = 120 J
2. F = W / d = 60 J / 2 m = 30 N
3. d = W / F = 350 J / 85 N = 4.12 m
4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts
5. W = P t = 500 W * 120 sec = 60,000 J
6. t = W / P = 550 J / 310 W = 1.77 sec
Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.