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lubasha [3.4K]
2 years ago
13

an elevator suspended by a vertical cable is moving downward but slowing down. the tension in the cable must be:

Physics
1 answer:
vladimir1956 [14]2 years ago
6 0

Answer:

F = M a is the vector equation involved

F = T - M g      are the forces acting on the elevator   (scalar equation)

T - M g = M a

T = M (a + g)    remember this a scalar

If a is slowing down then it must have a positive acceleration upwards

Therefor the tension in the cable must be greater than zero

When the tension increases to M g, a has increased to zero

For a to be zero, no acceleration, T = M g

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

8 0
1 year ago
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