A train leaves Houston and travels north at 65km/h. After it travels 150km, it turns left and travels 60km towards the west at 55km/h.

What is the total distance that the train traveled?

Answers

Answer 1

Answer:

210km

Explanation:

Given parameters:

Distance traveled northward  = 150km

Distance traveled westward  = 60km

Unknown:

Total distance traveled by the train  = ?

Solution:

Distance is the total length of path traveled by a body.

  Total distance  = distance traveled northward + distance traveled westward

Total distance  = 150km + 60km  = 210km

Answer 2

Answer:

210 km

Explanation:

Traveled 150 km north

Traveled 60 km west

150 + 60 = 210 km for distance traveled


Related Questions

asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *

Answers

Answer:

The same as

Explanation:

First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.

The angular velocity of a merry go round, in this instance, is given by

w = Δ/Δt

The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.

Transferring this further, the angular acceleration is given as

α = Δw/Δt

Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration

which property of potential energy distinguishes it from kinetic energy

Answers

Answer:

Shape and position

Explanation:

Hope this helps! :)

What type of numbers in a measurement are always significant?

Answers

Answer:

Non-zero digits

Explanation:

umm (ꏿ﹏ꏿ;)

Answer:

There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.

I hope this answered your question, have a nice day!

Need help ASAP please help

Answers

Answer:

.....

Explanation:

Why is struggling an important part of the process of learning?

Answers

Answer:

Students need a safe environment to take risks and struggle. It’s uncomfortable to struggle, but struggling—falling down and getting back up—is an important facet to learning. Productive struggle is not about being in pain or becoming frustrated.

Please mark branliest!

Explanation: it shows that your really trying your best and when you mess up its telling you that your actually learning from your mistakes.

1500 POINTS TO WHOEVER ANSWERS THIS FIRST!!!!!!!!!!

1. You drop a bowling ball 15m from the top of the science building. What is its
velocity as it impacts the ground?

2. A cheetah accelerates from 0 to 13m/s in 3 seconds. How far does it travel
during that time?

3. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. How high does the bullet go (neglecting air resistance)?

4. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. What is the bullets velocity when it falls back down and hits your
friend?

Answers

Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

Explanation:

During the slowing down process, there is a time when the angular speed is 0.5 rev/s. The fan arm is 0.8 m long. What is the speed of a point on the end of the fan arm at this time

Answers

Answer:

2.51 m/s

Explanation:

Given that,

Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]

Length of the fan arm, r = 0.8 m

We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :

[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]

So, the required speed is 2.51 m/s.

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