Answer:
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Explanation:
Answer:
i. + 22.5 m ii. 4.0 m
Explanation:
i. Image distance
Using the lens formula
1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown
So, we find v
1/v = 1/f - 1/u
= 1/+18 - 1/+90
= (5 - 1)/90
= 4/90
v = 90/4
= + 22.5 m
So the image is real and formed 22.5 m away on the other side of the lens.
ii Length of Shark
Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.
m = image distance/object distance
= v/u
= +22.5/+90
= 0.25
0.25 = image height/object height
So,
object height = image height/0.25
= 1.0 m/0.25
= 4.0 m
So, the length of the shark is 4.0 m
Answer:
28.3 m/s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 30°
Maximum height (H) = 10 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
Thus, we can obtain the minimum velocity cannon ball by using the following formula:
H = u²Sine² θ / 2g
10 = u² × (Sine 30)² / 2× 10
10 = u² × (0.5)² / 20
10 = u² × 0.25 / 20
10 = u² × 0.0125
Divide both side by 0.0125
u² = 10/ 0.0125
u² = 800
Take the square root of both side
u = √800
u = 28.3 m/s
Therefore, the minimum speed of the cannon ball is 28.3 m/s
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