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Lady bird [3.3K]

3 years ago

9

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

(and quite modest) acceleration. A train travels through a congested part of town at 7.0 m/s . Once free of this area, it speeds up to 13 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16s to reach a higher cruising speed.
Required:
What is the final speed?

1 answer:

Nataly [62]3 years ago

4 0

Answer:

v = 25 m / s

Explanation:

For this exercise we use the relations and kinematics

first part, the train accelerates from v₀ = 7.0 m/s to 13 m/s in a time t = 8.0 s

v = v₀ + a t

a = $\frac{v- v_o}{t}$

a = $\frac{13-7}{8}$

a = 0.75 m / s²

second part. Accelerate again for t = 16 s

v = v₁ + a t

for this interval the initial velocity is v₁ = 13 m / s

v = 13 + 0.75 16

v = 25 m / s

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An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal

german

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

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Momentum is a measure of the<br> of an object.<br> Which term accurately completes the sentence?

RideAnS [48]

Answer:

Momentum is define as the product of the mass and velocity of a body. It is measured in Kgm/s.

Explanation:

Momentum is the product of mass and velocity of an object. When an object or a body of mass 'm' is moving with velocity 'v', then its momentum can be determined as;

momentum (P) = mass × velocity

i.e P = m × v

= mv

It is measured in Kgm/s.

The change in momentum of a body is referred to as its impulse (Ft).

ΔP = m(v - u) = Ft

Where: P is the momentum of the object, m is its mass, v is its final velocity, u is the initial velocity, F is the force and t is the time in which the force acts.

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On James’s MP3 player, he has 12 sad songs and 40 upbeat songs that he wants to put into playlists. He wants to have the same nu

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Answer:

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Explanation:

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An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field

lawyer [7]

Answer:

a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

Em₀ = U = e DV

final point. Where they hit the target

Em_f = K = ½ m v2

energy is conserved

Em₀ = Em_f

e ΔV = ½ m v²

ΔV = $\frac{1}{2}$ mv²/e (1)

If the speed of light is c and this is 100% then 1% is

v = 1% c = c / 100

v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

$K_e = K_p$

K_p = ½ m v_e²

K_p = $\frac{1}{2}$ 9.1 10⁻³¹ (3 10⁶)²

K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

ΔV = Kp / M

ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

e ΔV = ½ M v²

v² = 2 e ΔV / M

v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

v² = 49,035 10⁸

v = 7 10⁴ m / s

Relation

v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

v/c= 2.33 10⁻⁴

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

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$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

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$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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