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3 years ago

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The owner of a van installs a rear-window lens that has a focal length of -0.304 m. When the owner looks out through the lens at

a person standing directly behind the van, the person appears to be just 0.237 m from the back of the van, and appears to be 0.343 m tall. (a) How far from the van is the person actually standing

1 answer:

Naddika [18.5K]3 years ago

7 0

Answer:

$p =-1.03$

Explanation:

From the question we are told that:

Focal length of lens $f=-0.304 m$

Image distance $q=0.237 m$

Height of image $H_i=0.343$

Generally the lens equation is mathematically given by

$\frac{1}{f} = \frac{1}{q} - \frac{1}{p}$

Where $p$ is Subject

$p = \frac{(qf) }{(f - q)}$

$p = \frac{(-0.237)(-0.304)) }{((-0.304) - (0.237))}$

$p =-1.03$

Therefore the distance between the person and the car is

$p =-1.03$

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

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Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

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Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

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For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

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$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

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We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

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