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SSSSS [86.1K]
3 years ago
10

An elevator is accelerating upward at a rate of 3.6 m/s2. a block of mass 24 kg hangs by a low-mass rope from the ceiling, and a

nother block of mass 90 kg hangs by a low-mass rope from the upper block. (a) what are the tensions in the upper and lower ropes?
Physics
1 answer:
kozerog [31]3 years ago
5 0
The answer:

<span>When the elevator accelerates upward at a rate of 3.6 m/s², the value of the acceleration becomes

</span>A=g+3.6=13.4 m/s²
and by using the newton's law, F=mass x A, we have 
T1= (24 + 90 )x 13.4= 1527.6 N, where T1 is the <span>Tension in upper rope
</span> and 
T2= ( 90 )x 13.4= 1206N, where T2 is the Tension in lower rope

When the elevator accelerates downward at a rate of 3.6 m/s², the value of the acceleration becomes
A=9.8 - 3.6 = 6.2 m/s²

T1= (24 + 90 )x 6.2= 706.8 N, where T1 is the Tension in upper rope
 and 
T2= ( 90 )x 6.2= 558N, where T2 is the Tension in lower rope


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Convert to metres

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Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the
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Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }

where

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S_B indicates the speed of the ship B

and

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now we will Substitute 0.40c for A and 0.60 for B in the equation which is

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3 years ago
What is the numerical value for standard atmospheric pressure?
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3 years ago
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
vredina [299]

Answer:

a) v_1=5.5172\ m.s^{-1}

b) v_2=-1.5152\ m.s^{-1}

c) v_3=4.6154\ m.s^{-1}

d) v_{avg}=3.8462\ m.s^{-1}

Explanation:

Given:

  • distance down the field in the first interval, d_1=16\ m
  • time duration of the first interval, t_1=2.9\ s
  • distance down the field in the second interval, d_2=-2.5\ m
  • time duration of the second interval, t_2=1.65\ s
  • distance down the field in the third interval, d_3=24\ m
  • time duration of the third interval, t_3=5.2\ s

a)

velocity in the first interval:

v_1=\frac{d_1}{t_1}

v_1=\frac{16}{2.9}

v_1=5.5172\ m.s^{-1}

b)

velocity in the second interval:

v_2=\frac{d_2}{t_2}

v_2=\frac{-2.5}{1.65}

v_2=-1.5152\ m.s^{-1}

c)

velocity in the third interval:

v_3=\frac{d_3}{t_3}

v_3=\frac{24}{5.2}

v_3=4.6154\ m.s^{-1}

d)

We know that the average velocity is given as the total displacement per unit time.

v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}

v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}

v_{avg}=3.8462\ m.s^{-1}

3 0
3 years ago
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