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2 years ago

13

A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s

peed of the roller coast? Is the velocity constant throughout the trip? Explain

2 answers:

Svetllana [295]2 years ago

6 0

All you have to do is divide 3000 by 100, Its 30

mina [271]2 years ago

6 0

<u>Answer:</u> The average speed of the roller coast is 30 m/s

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the runner, we use the equation:

$\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}$

We are given:

Total distance traveled by roller coast = 3000 m

Total time taken = 100 seconds

Putting values in above equation, we get:

$\text{Average speed of roller coast}=\frac{3000m}{100s}=30m/s$

Hence, the average speed of the roller coast is 30 m/s

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3 years ago

Read 2 more answers

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

3 0

2 years ago

If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

3 years ago