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Sophie [7]
4 years ago
9

A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b

reaking strength of 40.N. Calculate the maximum speed the cart can attain without breaking the string
Physics
1 answer:
luda_lava [24]4 years ago
8 0
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
Therefore the maximum speed the toy cart should have should be less than 20m/s²
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A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?
Allushta [10]
Hello,


Your answer to this problem is 400/3


Hope this helps!
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If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is
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Answer:

3

Explanation:

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3 years ago
The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of
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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  =4.51m/s  

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

        Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

  Now to obtain the initial energy stored

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Where x  is the length the spring is displaced

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         U = \frac{1}{2} * 627 * (0.6)^2

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   Now referring to the formula above

i.e          Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

                \frac{1}{2} mv^2 = 112.86 - \mu_kmgx

                v^2 = \frac{2(112.86 -\mu_kmgx)}{m}

                v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity = 9.8m/s^2  this displacement length of spring = 0.6

  Therefore   v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }

                        =4.51m/s      

           

8 0
3 years ago
Which of the following describes the relationship between the velocity and
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Explanation:

I think its a option pascal's principal

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