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3 years ago

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A man weighs 650 N. Inside a moving elevator, the floor exerts an upward force of 620 N on the man. What is the magnitude and di

rection of the man's acceleration?

2 answers:

nikitadnepr [17]3 years ago

6 0

Given that,

Weight of man, W= 650 N

Upward force by elevator, F= 620 N

a = ?

Since,

W = mg

m= W/g =650/9.81

m = 66.25 kg

Since,

F = ma

a = F/m

a = 620/66.25

a = 9.35 m/s²

Acceleration will be 9.35 m/s².

Since the force is in upward direction, it will accelerate the body in upward direction.

cestrela7 [59]3 years ago

3 0

Answer:Man's acceleration is 0.452 $\frac{m}{s^{2}}$ in downward direction

Explanation:

<em>Weight of man, W= 650 N</em>

<em>Normal force exerted by floor of elevator , N= 620 N</em>

<em>Since weight , W = mg</em>

<em>mass m= $\frac{W}{g}=\frac{650}{9.8}kg$ </em>

<em>Using Newtons 2nd law</em>

<em>W-N= ma where a= acceleration of man in the downward direction</em>

=> 650-620= $\frac{650}{9.8}kg\times a$

=> a= 0.452 $\frac{m}{s^{2}}$ <em>in downward direction </em>

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A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

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Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

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We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

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$A_1 v_1 = A_2 v_2$

where

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$r_2=\frac{0.618}{2}=0.309 m$ is the radius

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$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

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And the velocity in the constricted section is

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