All categories

3 years ago

If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?

Physics

1 answer:

Olenka [21]3 years ago

5 0

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

$v=\dfrac{d}{t}$

$d=v\times t$

$d=55.55\times 1800$

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

You might be interested in

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

guajiro [1.7K]

Answer:

$f_{e}$ = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

M = M₀ $m_{e}$

M = - L/f₀ 25/ $f_{e}$

Where f₀ and $f_{e}$ are the focal lengths of the lens and eyepiece, respectively, all values in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ( $f_{e}$ )

$f_{e}$ = - L / f₀ 25 / M

Let's calculate

$f_{e}$ = - 16 / 0.6 25 / (-400)

$f_{e}$ = 1.67 cm

The minus sign in the magnification is because the image is inverted.

$f_{e}$ = 1.7 cm

6 0

3 years ago

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

7 0

3 years ago

A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra

Sonja [21]

Answer:

Intensity, $I=1.101\ W/m^2$

Explanation:

Power of the light bulb, P = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

$I=\dfrac{40\ W}{4\pi (1.7\ m)^2}$

$I=1.101\ W/m^2$

So, the intensity of light is $1.101\ W/m^2$ .

6 0

4 years ago

In the circuit seen here, the resistor has a resistance of 3 ohms. If no change in the battery size occurs, what will happen to

Bumek [7]

The current will decrease as the resistance has now increased, meaning less current will be 'let through' the resistor. (assuming it's in series, there's no image)

6 0

4 years ago

Read 2 more answers

Liquid water heats up and cools down slowly. What causes this property?

Alja [10]

Answer:

water forms hydrogen bonds

Explanation:

3 0

3 years ago