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aleksandr82 [10.1K]

2 years ago

10

A train traveled 100 kilometers north in 2 hours. After exchanging some railroad cars, it traveled 100 kilometers south in 2 hou

rs along the same track. Why has the train's velocity changed

1 answer:

hodyreva [135]2 years ago

7 0

Answer:Due to change in direction

Explanation:

Given

Initially train has traveled a 100 km in North and after exchanging some railroad cars, it traveled 100 in south.

The velocity of the train changes as it direction of motion changes. Velocity is the vector quantity which require direction and magnitude for its reperesentation.

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A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two

Andrew [12]

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N

4 0

2 years ago

A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta

stepan [7]

Answer:

The value is $n = 2.347 *10^{17} \ photons$

Explanation:

From the question we are told that

The amount of power delivered is $P = 4.20 \ M W = 4.20 *10^{6} \ W$

The time taken is $t = 16.0ns = 16.0 *10^{-9} \ s$

The wavelength is $\lambda = 694.3 \ nm = 694.3 *10^{-9} \ m$

Generally the energy delivered is mathematically represented as

$E = P * t = \frac{n * h * c }{\lambda }$

Where $h$ is the Planck's constant with value $h = 6.262 *10^{-34} \ J \cdot s$

c is the speed of light with value $c = 3.0*10^{8} \ m/s$

So

$4.20 *10^{6} * 16*10^{-9}= \frac{n * 6.626 *10^{-34} * 3.0*10^{8} }{694.3 *10^{-9}}$

=> $n = 2.347 *10^{17} \ photons$

4 0

3 years ago

Withe light becomes a spectrum of seven colored lights after passing through a prism

Fiesta28 [93]

True. The 7 colors, also called a rainbow, are red, orange, yellow, green, blue, indigo, and violet. This rainbow is formed because the prism bends the white light and spreads it out into the colors it was made of. If there is more you were looking for, comment here.

8 0

3 years ago

Read 2 more answers

Una flecha tiene una rapidez de lanzamiento inicial de 18 m/s. Si debe dar en un blanco a 31 m de distancia, que está a la misma

maks197457 [2]

Answer:

34.8 and 55.2º

Explanation:

This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation

The scope equation is

R = v₀² sin 2θ /g

sin 2 θ = R g / v₀²

sin 2 θ = 31 9.8 / 18²

2 θ = sin⁻¹ 0.93765

θ = 34.8º

At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º

4 0

3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

3 years ago