All categories

2 years ago

What distance is required for a train to stop when traveling 50 miles per hour? half mile one and a half miles four miles

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

4 0

One and a half miles of distance is required for a train to stop when traveling 50 miles per hour.

Rate of change in position, or speed, is equal to distance traveled divided by time. To solve for time, divide the distance traveled by the rate.

Rearranging the formula,

Speed = distance / time.

Distance = speed × time.

Time = distance / speed.

Hence,

Distance = Speed × Time

⇒ Distance = 50 × 60 / 2000

⇒ Distance = 1.5 miles

⇒ Thus, the distance which is required to stop a train is 1.5 miles, which is, One and a half miles.

To learn more about Distance here

brainly.com/question/12319416

#SPJ4

You might be interested in

IS it necessary to place a lens after SLM?

Tamiku [17]

Answer:

a lens shouldn´t be necessary at all. What you are sending to the SLM should be a phase plate encoding Bessel beam you want.

hope it helpful for you

<h3>XxinstaprincixX</h3>

Explanation:

kese h o

7 0

3 years ago

Read 2 more answers

the sun on average is 93 billion miles from earth.how many meters is this? express it using power of ten and using a metric pref

Aleksandr [31]

Answer:

1.4966862 x 10^14 meters

1.55 x 10^11 km

3 0

3 years ago

The planet earth orbits around the sun and also spins around its own axis. True or False

posledela

Answer:

This is True

Explanation: The Earth rotates around its own axis, which results in day changing to night and back again. The Earth actually revolves around, or orbits, the sun. One revolution around the sun takes the Earth about 365 days, or one year. Forces at work in the solar system keep the Earth, as well as the other planets, locked into predictable orbits around the sun.

5 0

3 years ago

The number of molecules in a container is tripled and the kelvin temperature doubled. the volume remains unchanged. the new pres

Nonamiya [84]

N2 = 3*n1
T2 = 2*T1
V1 = V2

(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1

Since P2 is 6P1, it is 6 times greater than original pressure

3 0

4 years ago

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

seropon [69]

Answer:

The angular frequencies of all the 3 pendulums shall be same.

Explanation:

The time period of a simple pendulum with the approximation $sin(\theta )\approx \theta$ is given by:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is given by

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

As we can see that the angular frequency is independent on the initial angle (valid strictly for small angle approximations) we conclude that the angular frequencies of the 3 pendulums are the same.

6 0

4 years ago