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Vera_Pavlovna [14]
2 years ago
12

What do signal trainees do

Physics
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

A signal apprentice works for a rail company, learning the duties of a signal maintainer through on-the-job training and experience. As part of this apprenticeship, you work on a railway to assist a seasoned signal maintainer as they install, repair, inspect, and test signal equipment.

Explanation:

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I need to fill the gaps
andriy [413]

Answer:

1. combustion

2. oxygen

3. CO²

5 0
2 years ago
Which of the following changes will always increase the efficiency of a thermodynamic engine? Choose all correct statements.
Ilya [14]

Answer:B,C,D

Explanation:

Thermodynamic  efficiency is given by

\eta =1-\frac{T_C}{T-H}

\eta efficiency can be increased by Keeping _c constant and increasing T_H

Keeping T_H constant and decreasing T_c

by increasing \Delta T=T_H-T_c

by decreasing \frac{T_C}{T_H} ratio          

5 0
3 years ago
Read 2 more answers
A diver jumps up off a pier at an angle of 25° with an initial velocity of 3.2 m/s. How far from the pier will the diver hit the
rjkz [21]

Answer:

0.8 meters.

I just answered this ame question myself on a test I was taking.

6 0
2 years ago
Read 2 more answers
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
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