Answer:
you should provide the options for us to answer?
Answer:
See explanation below
Explanation:
In this case, you want to know if you put an object between these forces, which direction would go.
To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:
M = F * d (1)
And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:
anticlockwise moment = clockwise moment.
The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.
Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns
Anticlockwise moment = 34 * 3.5 = 119 Ns.
As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.
Hope this helps
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
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