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stepladder [879]

3 years ago

5

For which one of the following situations will the path length equal the magnitude of the displacement? a A ball on the end of a

string is moving in a vertical circle. b A toy train is traveling around a circular track. c A train travels 5 miles east before it stops. It then travels 2 miles west. d A ball is rolling down an inclined plane. e A ball rises and falls after being thrown straight up from the earth's surface.

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

d A ball is rolling down an inclined plane.

Explanation:

When path length is equal to the displacement

then we can say that the motion of the object must be in straight line so that the distance and displacement must be same

SO here we can say

a A ball on the end of a string is moving in a vertical circle.

In circular path distance and displacement is not same

b A toy train is traveling around a circular track.

In circular path distance and displacement is not same

c A train travels 5 miles east before it stops. It then travels 2 miles west.

Net displacement is 3 miles East while distance is 7 miles

d A ball is rolling down an inclined plane.

Here its motion is in straight line so we can say that path length and displacement will be same

e A ball rises and falls after being thrown straight up from the earth's surface.

In this type of to and fro motion path length is not same as displacement

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In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

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$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

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$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

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$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

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Read 2 more answers

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

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