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uysha [10]
2 years ago
15

A ledge on a building is 20 m above the ground. A taut rope attached to a 4.0 kg can of paint sitting on the ledge passes up ove

r a pulley and straight down to a 3.0 kg can of nails on the ground. If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?
Physics
1 answer:
Andrews [41]2 years ago
8 0

Answer:

 t = 5.4 s

Explanation:

from the question we are given :

height (s) = 20 m

mass of paint (Mp) = 4 kg

mass of nails (Mn) = 3 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

  • The net force accelerating the can of paint should be equal to the difference in weight of the can of paint and the can of nails.

            weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N

            weight of paint = mass nails x g = 4 x 9.8 = 39.2 N

             net force = 39.2 - 29.4 = 9.8 N

  • net force = total mass x acceleration

             9.8 = (3 +4) x a

              a = 1.4 m/s^{2}

  • from S = Ut + 0.5at^{2}  we can get  the time the carpenter has to catch the nails

          where U is the initial velocity and is 0 since the can was initially at            

            rest

           20 = (0 x t) + (0.5 x 1.4 x t^{2})

            20 = 0.7 x t^{2}

             t^{2} = 28.6

             t = 5.4 s

                   

         

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2 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?
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Answer:

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

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

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

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v_2=0.301\ m/s

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

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A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?
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(b) -50 m/sec

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(b) Time t = 10 sec

Now according to first equation of motion

So final velocity v = u-gt = 50-10×10 =-50 m/sec

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