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Tema [17]
2 years ago
10

An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist

ance, nor is there any friction between the elevator and the walls of the elevator shaft. Why does the upward force exerted on the elevator by the cable have a larger magnitude as the downward force of gravity on the elevator
Physics
2 answers:
shtirl [24]2 years ago
6 0

Answer:

F = W + ma  a> 0

Explanation:

For this exercise let's use Newton's second law

 we assume the upward direction as positive

          F - W = m a

          F = W + ma

          F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

               

We can see that since a> 0 the force F must have greater than the weight of the elevator

Anvisha [2.4K]2 years ago
6 0

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • At any time, there are two forces acting on the elevator: the tension in the cable T (upward) and the force of gravity (downward).
  • Taking the upward direction as positive, if we apply Newton's 2nd law, we will have the following equation:

       T - m*g = m*a (1)

  • Now, we know that the elevator is being lowered at a decreasing speed, which means that the acceleration must have an opposite direction to the displacement.
  • Since the displacement is downward, that means that the acceleration must be positive.
  • If a ≥ 0, this means that T- mg ≥ 0, so the tension T must have a larger magnitude as the downward force of gravity on the elevator.
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A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5
Phoenix [80]

Answer:

The propulsion force at the moment of departure, is 49 N  

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N      

Therefore, the propulsion force at the moment of departure, is 49 N  

7 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

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     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

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The solution for this problem is:

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