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dimaraw [331]
4 years ago
11

PLS HELP ASAP!!!

Physics
2 answers:
vitfil [10]4 years ago
8 0
N= 100 would be your answer
AveGali [126]4 years ago
6 0

Answer:

4=100

Explanation:

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A wave's _______________ and its frequency have a reciprocal relationship.
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Wavelength and frequency have a reciprocal relationship. If one doubles, the other halves.
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A passenger travels from the first floor to the 60th floor, a distance of 210 m in 35 s. what is the elevators speed
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  • Distance=210m
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\\ \tt\bull\leadsto Speed=\dfrac{Distance}{Time}

\\ \tt\bull\leadsto Speed=210/35=6m/s

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Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours
Nostrana [21]

Answer:

Approximately 116\; \text{miles} for the train from Boston to NYC Penn Station.

Approximately 105\; \text{miles} for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}.

\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}.

Calculate average speed of each train:

\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}.

\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}

Assume that it takes a time period of t for the trains to pass by each other after departure. Distance each train travelled would be:

s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t.

s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t.

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}.

Rearrange and solve for t:

(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}.

\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}.

Distance each train travelled in t = (539 / 279)\; \text{hour}:

\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}.

\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}.

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if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

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Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.

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