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3 years ago

Momentum of 0.5 kg moving ball with the velocity of 5 m / s

Physics

2 answers:

insens350 [35]3 years ago

6 0

Answer:

$\boxed{ \bold{ \huge{\boxed{ \sf{2.5 \: kg \: m \: / \: s}}}}}$

Explanation:

Mass ( m ) = 0.5 kg

Velocity ( v ) = 5 m / s

Momentum ( p ) = ?

Finding the momentum

$\boxed{ \sf{momentum \: ( \: p \: ) \: = \: mass \: \times \: velocity}}$

⇒ $\sf{momentum \: = \: 0.5 kg \times 5 \: meter \: \: per \: second}$

⇒ $\sf{momentum = 2.5 \: kg \: m \: / \: s}$

Hope I helped!

Best regards! :D

ankoles [38]3 years ago

3 0

Answer:

2.5

Explanation:

Momentum = Mass × Velocity

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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

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First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

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3 years ago

Momentum of 0.5 kg moving ball with the velocity of 5 m / s ​

Momentum of 0.5 kg moving ball with the velocity of 5 m / s