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GalinKa [24]
3 years ago
12

A 98-kg fullback, running at 5.0 m/s, attempts to dive directly across the goal line for a touchdown. Just as he reaches the lin

e, he is met head-on in midair by two 68-kg linebackers both moving in the direction opposite the fullback. One is moving at 2.0 m/s, the other at 4.0 m/s. They all become entangled as one mass.
(a) Sketch the event, identifying "before" and "after" situations. (Do this on paper. Your instructor may ask you to turn in this work.) This answer has not been graded yet.
(b) What is the velocity of the football players after the collision? (Take the positive direction to be the initial direction of the fullback.) m/s
(c) Does the fullback score a touchdown?
Yes
No
Physics
1 answer:
Troyanec [42]3 years ago
5 0

Answer:

(a) Explained below

(b) v_f=0.35\ m/s

(c) Yes

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total linear momentum of a system of particles or objects is conserved unless an external force is acting on the system. The formula for the momentum of a body with mass m and velocity v is P=mv. If there is a system of bodies, then the total linear momentum is the sum of the individual momentums

P=m_1v_1+m_2v_2+...+m_nv_n

When objects collide and join together, the only final mass is the sum of all masses, all traveling at the same speed.

Our m_1=98\ kg fullback runs at v_1=5\ m/s. Two two 68-kg linebackers attempt to stop him, one at -2.0 m/s and the other at -4.0 m/s. The negative value is because the run against the positive direction, taken in the direction of the fullback.

(a) Before the event, there is a total linear momentum, computed as the sum of the momentums of each player as shown

p_1=m_1v_1=(98)(5)=490 Kg\ m/s

p_2=m_2v_2=(68)(-4)=-272 kg\ m/s

p_3=m_3v_3=(68)(-2)=-136 kg\ m/s

p_t=p_1+p_2+p_3=390-272-136=82\ kg\ m/s

After the collision, all the players keep joined in one single mass of.

m_t=98+68+68=234\ kg

They will move at a speed which will be computed below

(b) The final momentum of the system is

p_f=m_tv_f=82\ kg\ m/s

Since the linear momentum is conserved, the final speed v_f is common to all of the players. Let's solve to find it

\displaystyle v_f=\frac{p_f}{m_t}

\displaystyle v_f=\frac{82}{234}

v_f=0.35\ m/s

(c) Since the final speed of the players is positive, it means the touchdown was actually scored, the fullback moved forward across the goal line, the positive reference.

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When an object is fully converted into energy the amount of energy liberated is
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Answer:

Mass, m = 4 kg

Explanation:

<u>Given the following data;</u>

Energy = 3.6 * 10^17 Joules

We know that the speed of light is equal to 3 * 10⁸ m/s.

To find the mass of the substance;

The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.

The equation illustrates, energy equals mass multiplied by the square of the speed of light.

Mathematically, the theory of special relativity is given by the formula;

E = mc^{2}

Where;

  • E is the energy possessed by a substance.
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Substituting into the formula, we have;

3.6 * 10^{17} = m * 300000000^{2}

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m = \frac {3.6 * 10^{17}}{9*10^{16}}

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A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?
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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
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To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

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Since S\neq   0 the heat engine is not reversible.

PART B)

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4 0
4 years ago
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

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Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

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