A. Diagram A
B. Diagram C & D
C. Diagram B
D. Diagram C & D
E. Diagram B
F. Diagram C & D
These are simplified representations of an object's body and the force vectors acting on it. Some of the main forces that are involve are normal force, friction, push or pull and gravity.
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
The net force is the vector
∑ F = (450 N) (cos(42°) i + sin(42°) j)
and two of the forces provided by the girls are
F₁ = (310 N) (cos(115°) i + sin(115°) j)
F₂ = (250 N) (cos(285°) i + sin(285°) j)
Then the force provided by the third girl is the vector
F₃ = ∑ F - F₁ - F₂
F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i
… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j
F₃ ≈ (400.722 N) i + (261.635 N) j
So, the third girl provided a force of magnitude
||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N
pointing in a direction
arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°
relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.
So, the third girl pushed with force 480 N [N57°E].
Answer:Both are correct
Explanation:
The opposite DTC also comprises of the wiring or ground. If the opposite DTC can be set it is the components that is faulty and if otherwise it is still the components that is faulty
Answer:
x₁ = 0.62 m
Explanation:
In this exercise the force is electric, given by Coulomb's law
F =
This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.
For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.
∑ F = F₁₂ - F₂₃
They ask us to find the point where the summaries of the force is zero.
F₁₂ - F₂₃ = 0
F₁₂ = F₂₃
let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere
k q₁q₂ / x² = k q₂q₃ / (d-x) ²
q₁ (d-x) ² = q₃ x²
let's solve
d² - 2 x d + x² = 
x²
x² (1 - ) - 2x d + d² = 0
we substitute the values
x² (1- 4/2) - 2 1.5 x + 1.5² = 0
x² (-1) - 3.0 x + 2.25 = 0
x² + 3 x - 2.25 = 0
let's solve the quadratic equation
x = [-3 ± 
] / 2
x = [-3 ± 4.24] / 2
x₁ = 0.62 m
x₂ = 3.62 m
since it indicates that the charge q₂ e places between the spheres, the correct solution is
x₁ = 0.62 m
