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Studentka2010 [4]
2 years ago
6

Ple can someone help me to answer this question

Physics
1 answer:
ra1l [238]2 years ago
8 0

The length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

<h3 /><h3>A sketch of the uniform stick and the mass</h3>

The sketch of the uniform stick and the mass will be two based on the given statement.

<h3>when the uniform stick balances on knife at 10 cm from one end;</h3>

-------------------------------------------------------  

↓      10cm           Δ       (L - 10 cm)         ↓

200g                                                      M

<h3>When the knife edge is moved 5 cm further</h3>

|---------------15 cm-----------------|

-----------------------------------------------------------------------

              ↓      8.75 cm           Δ       (L - 15 cm)         ↓

          200g                                                               M

From the first diagram, apply principle of moment;

200(10) = M(L - 10) ------- (1)

From the second diagram, apply principle of moment;

200(8.75) = M(L - 15)   ------- (2)

From equation (1); M = (2000) / (L - 10)

From equation (2); M = (1750) / (L - 15)

Solve (1) and (2);

(2000) / (L - 10) = (1750) / (L - 15)

1750(L - 10) = 2000(L - 15)

L - 10 = 2000/1750(L - 15)

L - 10 = 1.143(L - 15)

L - 10 = 1.143L - 17.14

17.14 - 10 = 1.143L - L

7.14 = 0.143L

L = 7.14/0.143

L = 50 cm

<h3>Mass of the uniform stick</h3>

M = (2000) / (50 - 10)

M = 50 g

Thus, the length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

Learn more about length of meter stick here: brainly.com/question/17225736

#SPJ1

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Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

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Mass of the basket ball, M = 594 g

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Now,

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u^2-u'^2 = 2as

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since it is free fall case

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a = g = acceleration due to gravity

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thus we have

u^2-0^2 = 2\times9.8\tiimes1.08

or

u = \sqrt{21.168}

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b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed (v_f) = v = 4.6 m/s

Then conservation of energy says  

\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2

substituting the values in the above equation we get

v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6

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or

v = -12.53m/s

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now the kinetic energy of the tennis ball

K.E = \frac{1}{2}mv^2

or

K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

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