Answer:
<em>Well, I think the best answer will be is </em><em>1.59 g/mL Good Luck!</em>
[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]
(a) 3,400 g x <u>0.001</u> = 3.40 kg [converting grams to kilograms]
(b) 220 cm x <u>0.01</u> = <u>2.2</u> m [converting centimeters to meters]
(c) 9.42 kg x <u>1000</u> = <u>9420</u> g [converting kilograms to grams]
(d) 6.53 m x <u>100</u> = <u>653</u> cm [converting meters to centimeters]
Answer:
answer is friction. MCQ A is answer
Answer:1). Distance of far point x=0.9m
Therefore, since the image is virtual
-f=-x = -0.9m
Power of the concave lenses = 1/f = 1/-0.9
= -1.11D
2 ) near point is 21cm = 0.21m
Power = 4-1/near point
= 4/0.21
= 14.2D.
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
