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kakasveta [241]
3 years ago
9

Ball B, with a mass of 30kg, is moving to the left at 10 m/s. With what velocity should ball A, with a mass of 10kg, move to the

right and collide with 2, so that Ball A rebounds with a velocity of 30 m/s, and Ball B with velocity of 10 m/s? Assume the collison to be perfectly elastic.
Physics
1 answer:
bogdanovich [222]3 years ago
6 0
Using the conservation of momentumwhere momentum = mass x velocity
so the let x be the initial velocity of the ball B

(30 kg) (-10 m/s) + (10 kg) (x) = (30 kg) (30 m/s) + ( 10 kg) (-10 m/s)
-300 + 10 x = 900 - 10010x = 900 - 100 + 30010x = 1100x = 1100 / 10x = 110 m/s
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Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two si
iVinArrow [24]

Answer: hello your question is incomplete below is the missing part

A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.

answer:

To 2 significant Figures = 1.6 m/s^2

Explanation:

<u>Calculate the magnitude of minimum acceleration at which the thief can descend </u>

we apply the relation below

Mg - T = Ma  --- ( 1 )

M = 69kg

g = 9.81

T = 58 * 9.81

a = ? ( magnitude of minimum acceleration)

From equation 1

a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69

  = 1.5639 m/s^2

To 2 significant Figures = 1.6 m/s^2

8 0
3 years ago
Una cámara fotográfica analógica (no digital) tiene dos lentes intercambiables. Uno de foco 55mm y el otro de 200 mm. Toma una f
Sergeu [11.5K]

Answer:

f = 55mm,     h ’= -9.89 cm

f = 200 mm,  h ’= 42.5 cm

Explanation:

For this exercise let's start by finding the distance to the image, using the equation of the constructor

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and image, respectively

lens with f₁ = 55mm = 0.55cm

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

         \frac{1}{q_1} = \frac{1}{0.55} - \frac{1}{10}

          \frac{1}{q_1} = 1.718

          q₁ = 0.582 m

lens with f₂ = 200mm = 2m

           \frac{1}{q_2} =   \frac{1}{2} - \frac{1}{10}

            \frac{1}{q_2} = 0.4

            q₂ = 2.5 m

the magnification of a lens is given by

            m = \frac{h'}{h} = -  \frac{q}{p}

             h ’= - \frac{q}{p} \ h

let's calculate for each lens

f = 55mm

             h '= - 0.582 / 10 1.7

             h ’= 0.0989 m

             h ’= -9.89 cm

f = 200 mm

             h '= - 2.5 / 10 1.7

             h ’= -0.425 m

             h ’= 42.5 cm

The negative sign indicates that the image is real and inverted

4 0
3 years ago
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Drupady [299]

Answer:

0.0084×91=0.7644

Explanation:

hope this helps

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Answer:

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-vesign shows the lens is <em><u>C</u></em><em><u>O</u></em><em><u>N</u></em><em><u>C</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em>

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cestrela7 [59]
B.??????? 
It's either B. or A.

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3 years ago
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