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ale4655 [162]
2 years ago
10

Rhea is running towards Sean and kicks a soccer ball to

Physics
1 answer:
baherus [9]2 years ago
3 0

Answer:

O The speed of the ball from Sean's reference frame would be greater than the speed of the ball from Rhea's reference frame.

Explanation:

Say the ball is kicked at 10 m/s and Rhea is running 3 m/s

Sean would see the ball approach at 10 m/s and Rhea approach at 3 m/s

Rhea would see Sean approach at 3 m/s and the ball depart at 10 - 3 = 7 m/s

The ball would see Sean approach at 10 m/s and Rhea depart at 7 m/s

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

7 0
2 years ago
How far could a rabbit run if it ran 36km/h for 5.0min?
Korolek [52]

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

4 0
2 years ago
Read 2 more answers
Fred wants to summarize mitosis in the cell cycle. Which statement describes mitosis?
salantis [7]
The statement that most accurately describes mitosis simply is <span>that mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself</span>. This is the most basic textbook definition, that is summary, of what the mitosis is.
6 0
2 years ago
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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
diamong [38]

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

S = 33.34Btu/R

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

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
2 years ago
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P
KIM [24]

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

E=\frac{F}{Q}

But the force F

F= \frac{kQ1Q2}{r^2}

But the electric field intensity due to a point charge Q at a distance r meters away is given by

E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }

<em>From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases</em>

6 0
2 years ago
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