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3 years ago

What happens to a shopping cart if you get it rolling and then release it?

Physics

1 answer:

Anit [1.1K]3 years ago

7 0

Answer:

It will slowly come to a stop as long as it is moving on a flat surface

Explanation:

If the cart is moving on a flat surface, it will slowly come to a stop due to the friction forces acting on its wheels.

If the cart is on an inclined surface, it may gain acceleration enough to overcome any friction force, and thus continue with an accelerated motion.

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a dog pulls on a pillow with a force of 8.4 N at an angle of 31 degrees above the horizontal. what is the x component of this fo

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4 years ago

Which type of device enables its users to see around corners?

Dmitriy789 [7]

Answer:

We can use optical fibers.

Explanation:

The optical fibers are the optical devices which is based on the total internal reflection.

The optical fiber has two layers, one is called core and the other is called cladding.

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As a ray of light falls on the core at more than the critical angle for that pair of media(core and cladding), it suffers the total internal reflection at the interface.

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3 years ago

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How to light prodeced by nature

frutty [35]

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3 years ago

Can somebody help me with this

Ivanshal [37]

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4 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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3 years ago