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tigry1 [53]
3 years ago
5

A car accelerates from rest at 4.7m/s^2. How much time does it need to attain a speed of 6m/s?

Physics
1 answer:
lesantik [10]3 years ago
8 0

Answer:

100mm per second =25 +29

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Why all traveller prefer to travel easily in the morning and night and not when the sun is hot​
eduard

Answer:

Travelers prefer to travel in morning because arthat time the temperature ️ is not too hot and they can easily travel here and there without any problem. and they also prefer to travel in night because moderate temperature ....

4 0
3 years ago
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0
3 years ago
Read 2 more answers
Reflection of the images example
NISA [10]

Answer:

reflection of water ,sound and water waves etc

8 0
3 years ago
A light with a second-order bright band forms a diffraction angle of 30.0°. The diffraction grating has 250.0 lines per mm. What
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I have no idea sorry
7 0
3 years ago
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Hydrogen line spectrum lies entirely within visible range
mart [117]

No, that's silly.

You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level.  All of those transitions ploop out
photons at Infrared wavelengths.

THEN next you get your Balmer series, where the electrons fall in
to the 2nd level.  Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.

And then there's the Lyman series, where electrons fall all the way
down to the #1 level.  Those are ALL in the ultraviolet. 
6 0
3 years ago
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