An areoplane initially at rest, undergoes a constant acceleration on of 2.5m/second down the runway for 30sec before taking off

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

4 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

$C=\dfrac{\varepsilon _oA}{d}$

$U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2$

If energy become double U' = 2 U then d'

$U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

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3 years ago

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English sorry that all English plz

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Distinguish between digital and analogue quantities with examples

valentinak56 [21]

Answer:

The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude.

Comparison chart.

Analog Digital

Example Human voice in air, analog electronic devices. Computers, CDs, DVDs, and other digital electronic devices.

Explanation:

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An areoplane initially at rest, undergoes a constant acceleration on of 2.5m/second down the runway for 30sec before taking off​

An areoplane initially at rest, undergoes a constant acceleration on of 2.5m/second down the runway for 30sec before taking off