Answer:
Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
Answer:
9 meter per second square
Explanation:
acceleration equation is final minus initial velocity divided by time in seconds.
36-15/3=9 meter per second square.
Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
Answer:
1.324 × 10⁷ m
Explanation:
The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.
Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.
We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have
Rω² = GME/R²
R(2π/T)² = GME/R²
R³ = GME(T/2π)²
R = ∛(GME)(T/2π)²
RE + h = ∛(GMET²/4π²)
h = ∛(GMET²/4π²) - RE
substituting the values of the variables, we have
h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m
h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m
h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m
h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m
h = 19.64 × 10⁶ m - 6.4 × 10⁶ m
h = 13.24 × 10⁶ m
h = 1.324 × 10⁷ m
