Answer:
They both tend to develop during the spring (March-June), reach peak intensity during the late autumn or winter (November-February), and then weaken during the spring or early summer (March-June)
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i = 
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>
Answer:
B
Explanation:
Two atoms which are isotopes of one another must have a different number of neutrons.
Isotopes are defined as atoms of the same element which have the same numbers of protons i.e. atomic number remains the same, but has different numbers of neutrons. It is observed that they have same chemical properties due to the same electronic configuration but physical properties differs.
<span>Using the kinematic equations below, we can calculate the initial velocity required.
Angle of projectile = 60 degrees
Acceleration due to gravity (Ay) = -10 m/s^2 (negative because downward)
Height of projectile (Dy) = 2m
Vfy^2=Voy^2 +2*Ay*Dy
Vfy = 0 m/s because the vertical velocity slows to zero at the height of its trajection.
So... 0 = Voy^2 + 2(-10)(2)
0 = Voy^2 - 40
40 = Voy^2
Sqrt40 = Voy
6.32 m/s = Voy
THIS IS NOT THE ANSWER. THIS IS JUST THE INITIAL VELOCITY IN THE Y DIRECTION.
Using trigonometry, Tan 60 = Voy/Vox. Tan 60 = 6.32/Vox. Vox*Tan 60 = Vox
Vox = 10.95 m/s. Now, using Vox = 10.95 and Voy = 6.32, we can use pythagorean theorem to find the total Vo. A^2 +B^2 = C^2
10.95^2 + 6.32^2 = C^2
Solving for C = 12.64 m/s
This is the velocity required to hit the surface. You can also calculate a bunch of other stuff now using the other kinematic equations.
V = 12.64 m/s</span>
Answer:
the pressure exerted on the turf when the player lands on his knee is 240,000 N/m².
Explanation:
Given;
combined weight of the players, F = 2400 N
measurement of the player's knee = 0.1 m by 0.1 m
The area of the player's knee is calculated as;
A = 0.1 m x 0.1 m = 0.01 m²
The pressure exerted on the turf when the player lands on his knee is calculated as;
P = F / A
P = (2400 N) / (0.01 m²)
P = 240,000 N/m²
Therefore, the pressure exerted on the turf when the player lands on his knee is 240,000 N/m².
