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3 years ago

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A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe

ed of 5.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg

1 answer:

dezoksy [38]3 years ago

5 0

Answer:

the magnitude of the average contact force exerted on the leg is 3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F $_{hand$

using equation for impulse in momentum

F $_{hand$ × t = m( v - v₀ )

we substitute

F $_{hand$ × 0.00265 = 1.75( 0 - 5.25 )

F $_{hand$ × 0.00265 = 1.75( - 5.25 )

F $_{hand$ × 0.00265 = -9.1875

F $_{hand$ = -9.1875 / 0.00265

F $_{hand$ = -3466.98 N

Next we determine force on the leg F $_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F $_{leg$ = - F $_{hand$

we substitute

F $_{leg$ = - ( -3466.98 N )

F $_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N

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Bill is farsighted and has a near point located 125 cm from his eyes. Anne is also farsighted, but her near point is 75.0 cm fro

Arisa [49]

Answer:

a) The closest object Anne can see, while wearing Bill's glasses is at approximately 22.39 cm relative to her eyes

b) The closest object Bill can see while wearing Anne's glasses is at approximately 26.38 cm relative to his eyes

Explanation:

The point where Bill has a near point = 125 cm

The point where Anne has a near point = 75.0 cm

Their vision are both corrected to the normal near point = 25.0 cm

The distance of their glasses from the eye, d = 2.0 cm

The lens formula is presented as follows;

$\dfrac{1}{f} =\dfrac{1}{d_0} +\dfrac{1}{d_i}$

The required distance of the of the object from Bill's glass, $d_{oB}$ = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Bill's glass, $d_{iB}$ = -(125 cm - 2.0 cm) = -123 cm

Therefore, for Bill, we have;

$\dfrac{1}{f_B} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}$

Plugging in the values gives;

$\dfrac{1}{f_B} =\dfrac{1}{23} -\dfrac{1}{123} = \dfrac{100}{2,829}$

Therefore;

$f_B$ = 2,829/100 cm = 28.29 cm

The required distance of the of the object from Anne's glass, $d_{oA}$ = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Anne's glass, $d_{iA}$ = -(75 cm - 2 cm) = -73 cm

The focal length for Anne is therefore;

$\dfrac{1}{f_A} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}$

Plugging in the values gives;

$\dfrac{1}{f_A} =\dfrac{1}{23} -\dfrac{1}{73} = \dfrac{50}{1,679}$

Therefore, we have;

$f_A$ = 1,679/50 cm = 33.58 cm

a) When Anne wears Bill's lasses, we have;

$\dfrac{1}{f_B} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}$

Therefore, we get;

$\dfrac{1}{28.29} =\dfrac{1}{d_{0A}} -\dfrac{1}{73}$

$\dfrac{1}{d_{0A}} = \dfrac{1}{28.29} + \dfrac{1}{73} = \dfrac{10,129}{206,517}$

$d_{0A}$ ≈ 20.39 cm

The distance of the closest object Anne can see, from her eye, $d_{oe}$ = $d_{0A}$ + d

∴ $d_{oe}$ ≈ 20.39 cm + 2.0 cm = 22.39 cm

The distance of the closest object Anne can see, from her eye, $d_{oe}$ ≈ 22.39 cm

b) The closest object that can be seen when Bill wears Anne's glasses, we have;

$\dfrac{1}{f_A} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}$

Therefore;

$\dfrac{1}{33.58} =\dfrac{1}{d_{0B}} -\dfrac{1}{123}$

$\dfrac{1}{d_{0B}} = \dfrac{1}{33.58} +\dfrac{1}{123} = \dfrac{7,829}{206,517}$

∴ $d_{oB}$ ≈ 26.38 cm

The closest object Bill can see while wearing Anne's glasses, $d_{oB}$ ≈ 26.38 cm.

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