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3 years ago

A high speed train in japan travels a distance of 300 km in 3.60x10^3 seconds . What is the average speed of this train

Physics

1 answer:

Bezzdna [24]3 years ago

8 0

83.33km per second.

Speed = distance ÷ time.

Speed = 300 ÷ 3.60000

Speed = 83.33km per second

hope that helps.

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Perform the calculation and report your answer using sig figs. 657.70 - 26.543

Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that the number "0" is what is usually (but not always) affected while trying to identify significant figures. Hence, all other numbers/digits are always significant. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70 - 26.543 = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0

3 years ago

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

3 years ago

Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the

shutvik [7]

Period = 6 seconds and $frequency = 0.167Hz$ .

Explanation:

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of $\frac{1}{t}$ . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency = $\frac{1}{t}$

⇒ $frequency = \frac{1}{time}$

⇒ $frequency = \frac{1}{6}$

⇒ $frequency = 0.167Hz$

Therefore , Period = 6 seconds and $frequency = 0.167Hz$ .

7 0

3 years ago

If you can run at a speed of 8 miles per hour and you want to run to the store 16 miles away, how

MArishka [77]

Answer:

16÷8=2

Explanation:

if you run 8 mi an hour than in 16 mi you would have run 2 hours

3 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago