Answer: D. movement of material along a coast by waves that approach at an angle to the shore.
Explanation:
Longshore drift is also referred to as the littoral druft and it means the sediment that is moved by the longshore current.
Longshore drift is the movement of material along a coast by waves which approach at an angle to the shore but then recede down the beach.
Therefore, the correct option is D.
Answer:
The fastest object is the sphere, so it is the winner
Explanation:
To know which object will arrive faster down, let's look for the velocity of the center of mass of each object. Let's use the concept of mechanical energy
Highest point
Em₀ = U = mg y
Lowest point
= K = + = ½ I w² + ½ m ²
Angular velocity is related to linear velocity.
v = w r
w = v / r
= ½ I ²/r² + ½ m ²
= ½ (I / r² + m) ²
Energy is conserved
Em₀ =
mg y = ½ (I / r² + m) ²
= √2 g y / (I / mr² +1)
With this expression we can know which object arrives as a higher speed, therefore invests less time and is the winner. Let's calculate the speed of the center of mass of each
Ring
I = m r²
= √ (2 g y / (m r² / mr² + 1))
= √ (2gy 1/2)
= (√ 2gy) 0.707
Solid sphere
I = 2/5 m r²
= √ (2gy / (2/5 m r² / mr² + 1)
= √ (2gy / (7/5))
= √ (2gy 5/7)
= (√ 2gy) 0.845
Cylinder
I = ½ m r²
= √ (2gy / ½ mr² / mr² + 1)
= √ (2gy / (3/2))
= √ (2g y 2/3)
= (√ 2gy) 0.816
The fastest object is the sphere, so it is the winner when descending the ramp
Voltage = (current) x (resistance) (Ohm's law)
Voltage = (15 A) x (18 ohms)
<em>Voltage = 270 volts</em>
Answer:
u = 8.3 x 10⁻⁴ m/s
Volume to Pass = 5 x 10⁻⁴ L = 0.5 mL
Explanation:
First we find area of pipe:
A = πd²/4
A = π(2.54 x 10⁻² m)²/4
A = 0.02 m²
Now,
Q = (1 L/min)(1 min/60 s)(0.001 m³/1 L)
Q = 1.66 x 10⁻⁵ m³/s
Therefore,
u = Q/A
u = (1.66 x 10⁻⁵ m³/s)/(0.02 m²)
<u>u = 8.3 x 10⁻⁴ m/s</u>
Time for 10 mL:
T = 0.01 L/Q
T = (0.01 L)/(1 L/min)
T = (0.01 min)(60 s/1 min)
T = 0.6 s
Liters of water to pass:
Volume to Pass = uT
Volume to Pass = (8.3 x 10⁻⁴ m/s)(0.6 s)
<u>Volume to Pass = 5 x 10⁻⁴ L = 0.5 mL</u>