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stiks02 [169]
2 years ago
5

The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.

Physics
1 answer:
Doss [256]2 years ago
7 0

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

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Concept map for kinetic energy, work and power
Helen [10]

Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:

W=K_f - K_i

where

K_f is the  final kinetic energy

K_i is the initial kinetic energy

Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as

P=\frac{W}{t}

where

W is the work done

t is the time elapsed

Learn more about kinetic energy, work and power:

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#LearnwithBrainly

8 0
3 years ago
Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If
cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

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A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

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Stars fuse most of the hydrogen in the entire star before they die.<br> 2 poir<br> True<br> False
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Once all hydrogen is depleted the star begins to die
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