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4 years ago

A car goes from 80 km/h to 20 km/h in 0.5 h. What is the acceleration in km/h2?

Physics

2 answers:

Nikolay [14]4 years ago

8 0

The answer is attached below.
Let me know if you don't understand it and I'll explain even more for you ☺♥

masya89 [10]4 years ago

8 0

Alright! So a car goes from 80 km/h to 20 km/h in 0.5 h. They want to know the acceleration in km/h2.

To Find it

equalate it

a(vf–vI)/t = (20 km/h–80 km/h)/0.5 h = –120 km/h2

Hope this helps and have a wonderful day!

<em>Tobey</em>

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3 years ago

Which of the following is a small star that has reached the end of its stellar evolution?

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D.) White Dwarf

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3 0

3 years ago

Read 2 more answers

In an electrically neutral atom of any element, there are equal numbers of

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Answer:

B) Electrons and protons

Explanation:

Each proton has a positive charge, and each electron has a negative charge. If the atom is electrically neutral, there must be the same number of electrons and protons.

A) and C) are wrong. Neutrons have no charge, so they can't counteract that of the protons or the electrons.

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3 years ago

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}$ .

8 0

2 years ago