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1 year ago

6

A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 N and 15

10 N, respectively. The acceleration of the cable is 0.620 m/s2, upward. What is the tension in the cable(a) below the worker and(b) above the worker?

1 answer:

Alex777 [14]1 year ago

4 0

Answer:(a)2631.46 N

(b)1605.51 N

Explanation:

Assuming there is a cable between crate and worker

Given

weight of worker $W_w=965 N$

Weight of crate $W_c=1510 N$

mass of worker $m_w=\frac{965}{g}=98.46 kg$

mass of crate $m_c=\frac{1510}{g}=154.08 kg$

acceleration of system $a=0.620 m/s^2$

Tension in the wire carrying Worker

$T-(m_w+m_c)g=(m_w+m_c)a$

$T=(m_w+m_c)(g+a)$

$T=(252.54)(9.8+0.62)$

$T=2631.46 N$

(b)Tension in the string above

$T_2-m_c\cdot=m_ca$

$T_2=m_c(g+a)$

$T_2=154.08(9.8+0.62)$

$T_2=1605.51 N$

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(a).

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solving for $t$ we get:

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(b).

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1 year ago

An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

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where
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v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
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In the first part of the problem:
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If we plug the numbers into the formula, we find
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b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
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Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
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2 years ago

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Leno4ka [110]

Answer:

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sin 30 = T_{y} / T

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2 years ago

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Bezzdna [24]

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(b)
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2 years ago

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Gre4nikov [31]

Answer:

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Equating the difference of equation (2) and equation (1); we have:'

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7 0

2 years ago

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