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KATRIN_1 [288]
3 years ago
11

What is the momentum of a 0.2kg ball with a speed of 5 m/s

Physics
1 answer:
Eddi Din [679]3 years ago
3 0
Momentum

= mass x velocity

= 0.2 x 5

= 1 kg m/s
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Explanation:

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A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
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Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

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Without being able to look into the earth , a scientist would not be able to determine if this region has karst topography
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Answer: False

Explanation: They can see it by  karst topography!

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alekssr [168]

Explanation:

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(a) <u>The</u><u> </u><u>segment</u><u> </u>A shows acceleration as velocity increases with the increase in time.

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(b) <u>The</u><u> </u><u>segment</u><u> </u>C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.

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(d) The acceleration of segment B is <u>zero</u><u>.</u> As there in no change in curve and it is moving with uniform velocity.

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\:

<h2>Thank you!</h2>
3 0
2 years ago
What law of motion does this picture represent?​
enyata [817]
Which picture are you talking about?
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2 years ago
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