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skad [1K]
3 years ago
15

All objects, regardless of their mass, fall with the same rate of acceleration on

Physics
2 answers:
tamaranim1 [39]3 years ago
4 0

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

crimeas [40]3 years ago
3 0
True this is true because the acceleration is equal
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Assume a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will ________________ between Resist
kow [346]

If a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2 and then recombine at Point B

<h3>Parallel circuit</h3>

A  parallel circuit is a circuit with separate branches with a common endpoint. In a parallel circuit, the voltage across each branch is the same but the currents vary. The total current is the sum of the currents flowing through each component.

If a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2 and then recombine at Point B.

Find out more on Parallel circuit at: brainly.com/question/80537

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2 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

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