Well by using the equation Speed = Distance / Time.
So, if the speed is 1.35 m/s and the time is 0.4s, then 1.35 = distance / 0.4
by rearranging the equation we get, 1.35 x 0.4 = distance.
therefore the distance travelled is 0.54 metres = 54 cm.
The crossbar is 10 feet above the ground.
Use the sine ratio to find the height of the kite: sine = opposite (height of kite with the horizontal) divided by the hypotenuse (the string) Rearrange the formula: sine*hypotenuse = opposite sine 25 degrees*150 = 63.39273926 feet Height of kite above the ground: 63.39273926+4.5 = 67.89273926 feet Therefore the kite is 68 feet above the ground to the nearest foot
43.3
17.8 meters
1000000=M with horizontal bar above it 90000=XC with horizontal bar above it 5000=V with horizontal bar above it 4=IV Therefore: ..............._____ 1905005= MXCVIV
T= Time, h=height, g= gravity Formula T=Sqr.rt (2h/g) Given= T=.350s g=9.8 1. .350=Sqr.rt(2h/9.8) (raise both sides to the second power) 2. .350^2=Sqr.rt(2h/9.8)^2 ( this will equal to) 3. .1225=2h/9.8 ( multiply both sides by 9.8) 4. 1.2005=2h (divide both sides by 2) 5. .60025=h (answer)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
The horizontal line above vowels means the letter is a long vowel.
900000=CM with horizontal bar above it 90000=XC with horizontal bar above it 9000=MX with horizontal bar above X only 900=CM Therefore: ............_____..._ 999900=CMXCMXCM
A Horizontal Line
600000=DC with horizontal bar above it 50000=L with horizontal bar above it 4000=IV with horizontal bar above it 100=C 20=XX 3=III therefore .............______ 654123=DCLIVCXXIII