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olasank [31]
3 years ago
10

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai

nter is standing 2 m from the first rope and 8 m from the second rope. What is the tension in the first rope (in Newtons)
Physics
1 answer:
RSB [31]3 years ago
8 0

Explanation:

For equilibrium, \sum M = 0.

So,   8 m \times mg - (10 m) T_{1} = 0

             T_{1} = \frac{8 \times mg}{10}

                        = \frac{8 \times 90 \times 9.8}{10}

                        = 705.6 N

Also, for equilibrium \sum F_{y} = 0

              T_{1} + T_{2} - mg = 0

or,         T_{2} = mg - T_{1}

                        = 90 \times 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

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Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the
WINSTONCH [101]

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

t_A=0.55s

t_B=0.45s

(a) The distance from the kicker to each of the 2 spectators is given by:

d_A=v \times t_A

where,

v= speed of sound

t_A=time taken for the sound waves to reach the ears

d_A=343\times 0.55=188.65m

(b)d_B=v \times t_B

where,

v= speed of sound

t_B=time taken for the sound waves to reach the ears

d_B=343\times 0.45=154.35m

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m

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another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

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