A 600 kg elevator moves with a downward acceleration of 1.90 m/s^2. What is the

Question

A 600 kg elevator moves with a downward acceleration of 1.90 m/s^2. What is the tension in cable that supports the elevator? 1.75 kN 2.75 kN 3.75 kN 4.75 kN Write the correct dimensional form for the following using mass; M, length: L, and time: T: point each) Displacement Acceleration Velocity Centripetal Acceleration Distance C^rightarrow A^rightarrow + B^rightarrow will have the maximum magnitude when A^rightarrow and B^rightarrow are perpendicular to each other The angle between A^rightarrow and B^rightarrow is 90 degree. The angle between A^rightarrow and B^rightarrow is 0 degree. The angle between A^rightarrow and B^rightarrow is 180 degree. The angle between A^rightarrow and B^rightarrow is 45 degree. None of these You accidentally knock a flowerpot off the top of the balcony wall of your apartment, exactly 100 m above the ground. It passes by a window whose top is 60.0 m below the your balcony wall. If the window is 1.00 m high, how long does the flowerpot take to window? 0.214 s 0.142 s 0.412 s 0.124 s From the figure below, if the magnitude of v^rightarrow_tot is 7.0 m/s and the angles shown are theta_1 and theta_2 35 degree the magnitudes of v^rightarrow_A and v^rightarrow_B are:

Explanation / Answer

11)

mg - T = ma

So, 600*9.8 - T = 600*1.9

So, T = 4740 N = 4.74 kN

12)

a) Displacement = [ M^0 L T^0 ]

b) Acceleration = [ M^0LT^-2 ]

c) Velocity = [M^0 LT^-1]

d) Centripetal acceleration = [ M^0LT^-2 ]

e) distance = [M^0 L T^0]

13) Answer : c) angle between A and B is 0 deg

14)

s = ut + 0.5*at^2

So, 60 = 0 + 0.5*9.8*t^2

So, t = 3.5 s

After traveling 1m :

s = ut + 0.5*at^2

So, 61 = 0 + 0.5*9.8*t'^2

So, t' = 3.53 s

So, time taken to pass = 3.53 - 3.5 = 0.03 s

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