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Kaylis [27]
2 years ago
6

A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from

its ends respectively. find the distances from the centre of the rule at which a 2N weight must be suspend a
A=0.09m​
Physics
2 answers:
Rzqust [24]2 years ago
7 0

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

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

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

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Amiraneli [1.4K]2 years ago
5 0

Answer:

The answer is 29 cm and  43.5 cm

Explanation:

The loop  A is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark ,T at 70 cm mark , 2 N at x

Taking the sum of the torques about B:

By using formula:

∑τ = Iα

(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0

18 Ncm − 2 N (x − 70 cm) = 0

2 N (x − 70 cm) = 18 Ncm

x − 70 cm = 9 cm

x = 79 cm

therefore ,the distance from the center is |50 cm − 79 cm

                                                    = 29 cm.

(b) when loop B is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark ,T at 20 cm mark ,2 N at x

Taking the sum of the torques about A:

∑τ = Iα

(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0

-27 Ncm − 2 N (x − 20 cm) = 0

2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

Therefore the distance from the center is |50 cm − 6.5 cm

i.e. 43.5 cm

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Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second
Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light   λ  = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

=  2.09 /1.65  

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0
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Answer:

Explanation:

Given

mass of crane m=1400\ kg

distance moved d=40\ m

Since it is moving with a constant velocity therefore net force on it is zero

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Work done by Tension T is

W_T=T\cdot d

W_T=1400\times 9.8\cdot 40

W_T=548.8\ KJ

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

W_T+W_g=0

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A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?
AysviL [449]
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
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v' = 1.41*3 = 4.23
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So, option B is your answer!

Hope this helps!

7 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

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Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

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To find,

The speed of the large cart after collision.

Solution,

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v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

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So, the speed of the large cart after collision is 0.301 m/s.

4 0
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