Answer:
a) Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Explanation:
a) Statement A : 2.567km to two significant figures.
2.567km 2. S.F = 2.6km
Statement B : 2.567km to three significant figures.
2.567km 3 S.F = 2.57km
Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) statement A: (2.567 km + 3.146km) to 2 S.F
(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km
Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).
2.567km to 2 S.F = 2.6km
3.146km to 2 S.F = 3.1km
2.6km + 3.1km = 5.7km
Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Answer:
Radio waves are the lowest frequency
Explanation:
Here is the order of frequencies highest to lowest to help you in the future:
gamma rays
X-rays
ultraviolet radiation
visible light
infrared radiation
radio waves
Answer:
Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.
Here since both children and merry go round is our system and there is no torque acting on this system
So we will use angular momentum conservation in this
now here we have
Now when children come to the position of half radius
then we will have
now from above equation we have
