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BartSMP [9]
2 years ago
12

Just for fun, a person jumps from rest from the top of a tall cliff overlooking a lake. in falling through a distance h, she acq

uires a certain speed v. assuming free-fall conditions, how much farther must she fall in order to acquire a speed of 2v? express your answer in terms of h.
b. would the answer to part
a. be different if this event were to occur on another planet where the acceleration due to gravity had a value other than 9.80 m/s2? explain.
Physics
1 answer:
RideAnS [48]2 years ago
4 0
For this we need to use several formulas:
v = a*t
This simply means that after 1 second we will have some speed and if we double time, speed will double as well.

But, relation time-traveled distance isn't linear. Traveled distance we calculate:
s = 1/2*a*t^2                we can say that h=s
so if we double time, traveled distance will increase 4 times because of square relation between time and traveled distance.

New traveled distance (h2) will be: h2=4*h

b)  Because distance depends on gravity acceleration in free fall h2 will certainly change but the relation h2=4*h will remain the same.
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Answer:

The Sun has a north and south pole, just as the Earth does, and rotates on its axis. However, unlike Earth which rotates at all latitudes every 24 hours, the Sun rotates every 25 days at the equator and takes progressively longer to rotate at higher latitudes, up to 35 days at the poles. This is known as differential rotation.

Explanation:

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How is refraction different from reflection?
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<span>The angle of refraction is not necessarily equal to the angle
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What will decrease over time as a result of improved
bagirrra123 [75]

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C

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4 0
2 years ago
Read 2 more answers
What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
If an automobile battery is rated at 20 ampere-hours, what is the maximum current that can be supplied for 19 minutes?
Elenna [48]
20Ah is the number of charge that can be supplied at 20A for 1 hour. If you wish to drain it in 19 minutes, then the current is:
20*60 = A_{2}*19
A_{2}=63.2A


5 0
2 years ago
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