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3 years ago

6

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

its final position?

1 answer:

ycow [4]3 years ago

5 0

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

When in orbit, astronauts experience weightlessness what is this caused by?

erica [24]

no gravity or dark matter

4 0

2 years ago

Read 2 more answers

Find the mass of an object if a 40 N of force causes the object to accelerate at 5.5 m/s/s

Murrr4er [49]

Answer:

<h3>The answer is 8 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{40}{5} \\$

We have the final answer as

<h3>8 kg</h3>

Hope this helps you

7 0

3 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

igomit [66]

Answer:

Angular acceleration of the disk will be $\alpha =10.714rad/sec^2$

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius $r=\frac{d}{2}=\frac{0.3}{2}=0.15m$

Moment of inertia of disk is given by $I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2$

Force is given by F=4 N

Torque is given as $\tau =Fr=4\times 0.15=0.6N-m$

We also know that torque is given by $\tau =I\alpha$

$0.6=0.056\times \alpha$

$\alpha =10.714rad/sec^2$

5 0

2 years ago

Which geologic feature most likely be represented by contour lines drawn far apart from one another?

andrey2020 [161]

Answer;

A plain

Explanation;

A contour line connects points of the same elevation. Contour lines are usually curves. Closed contours represent hills.

Contour lines can not cross since they represent different elevations. A contour interval is the difference in elevation between one contour and an adjacent contour.

5 0

3 years ago