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2 years ago

Help please..................

Physics

1 answer:

Shtirlitz [24]2 years ago

7 0

<h3><em>If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart. If two objects with opposite charges are brought towards each other the force will be attractive, it will pull them towards each other.</em></h3><h3><em>hope it helps.... thank you....</em></h3>

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Photosynthesis and respiration are best described as

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2 years ago

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this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

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ANSWER:

(a) 1036 N

(b) -1036 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

$\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}$

(b)

$\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}$

(c)

$\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}$

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3 months ago

if you rub a balloon on your hair, it becomes charged, and as a result It can stick to your head without being held there. why d

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2 years ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

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3 years ago

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2 years ago