About 12 hours is the time between a morning high tide and the next high tide
Explanation:
The Earth’s rotation happens between two tidal bulges
The “periodic rise and fall” of the surface water levels of the ocean is called tides. The gravitational action and interaction on the earth by the sun and the moon causes these tides. Different regions of the World experiences different patterns of tides like the diurnal, semi-diurnal etc.
When there is one high and one low tide occurring on a lunar day, then it is diurnal pattern. Semi-diurnal pattern occurs when there are two equal high and low tides on a single lunar day.
Since the Earth’s rotation happens between two tidal “bulges” on each lunar day, the coastal areas can experience two high and two low tides in every 24 hours plus 50 minutes.
Accordingly the time between two high tides would be 12 hours plus 25 minutes. Similarly, the time gap between a high to low tide would be 6 hours plus 12.5 minutes.
What happens when the light hits the glass depends on what it was in before it hit the glass.
WHILE it's in the glass, the speed of light doesn't change.
Explanation:
The mass of a ball, m = 2 kg
It is traveling with a speed of 10 m/s
The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :
The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.
Hence, the required kinetic and potential energy is 100 J.
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
<h2>
Speed of ball after 2 seconds is 8.62 m/s downward.</h2>
Explanation:
Let upper direction be positive
We have equation of motion v = u + at
Initial velocity, u = 11 m/s
Final velocity, v = ?
Time, t = 2 s
Acceleration,a = -9.81 m/s²
Substituting
v = u + at
v = 11 + -9.81 x 2
v = -8.62 m/s
Speed of ball after 2 seconds is 8.62 m/s downward.
