Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
Answer:
O2 has two more electrons compared to N2, with extra 2 electrons in the higher energy anti-bonding orbitals known as Diradical. These electrons have higher energy and are unpaired; therefore, O2 is more reactive
Explanation:
At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.
<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
- At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
- However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
- u= initial velocity, θ= angle of projection
<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
- During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
- But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.
Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory
A- neither the ball's velocity nor its acceleration are zero.
B- the ball's acceleration points upward.
C- the ball's acceleration is zero but its velocity is not zero.
D- the ball's velocity points downward.
Learn more about the projectile motion here:
brainly.com/question/24216590
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Answer:
A- mass and type of material
B- type of material
C- Temperature
Explanation:
thx
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
