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andre [41]
3 years ago
6

A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli

es the brakes. Initially, the train is moving at 50 m/s and slows down at 2.5 m/s^2. If the train stops at the station, how far before the station did the engineer apply the brake?
Physics
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s

Time taken by the train to stop is 20 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m

∴ The engineer applied the brakes 500 m from the station

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Calculate the energy of a photon of electromagnetic radiation whose frequency is 2.94 x 10 to the 14th xbox one
DIA [1.3K]

Answer:

1.95\cdot 10^{-19} J

Explanation:

The energy of a photon is given by:

E=hf

where

h=6.63\cdot 10^{-34}Js is the Planck constant

f=2.94\cdot 10^{14} Hz is the frequency of the photon

Substituting the numbers into the equation, we find:

E=(6.63\cdot 10^{-34} Js)(2.94\cdot 10^{14} Hz)=1.95\cdot 10^{-19} J

3 0
2 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

5 0
3 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
What is the simplest way to build a graph of kinetic and potential energy?
Elena-2011 [213]

Answer:

analize the levels of kinetic and potential energy and look up a guide to graph it and follow that

Explanation:

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3 years ago
Why do skiers often wear sunglasses while they are skiing
yulyashka [42]
There is a lot of glare off of the ice, due to the sun and it also is always good to have eye protection in case you fall face first :P
4 0
3 years ago
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