Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:
hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:
hence, the mean velocity is 8.4 m/s
Greetings!
"<span>How many significant figures does 0.00340 have?"...
A significant figure is:
-A non zero number
-A zero in between non zero numbers
-Trailing zeros to the right of the decimal point
Therefore, this number has three significant figures:
3,4,0
Hope this helps.
-Benjamin</span>
Answer:
A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.
A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.
An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.
I HOPE THIS IS HELPFUL.
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ 
............1
put here value
= 1.75× × 
= 0.175 v
and
area between air and puck is given by
Area = 
area = 
area = 7.85 × 
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 × v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds
