Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

now we know that

so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
1.8 mol of Na. hope this helps
Answer:
Take the numbers and fill them in graph.
Plot time as X and speed as Y and afterward join then like a line graph.
Answer:
(a) 
(b) 
Explanation:
Represent losing with L and winning with W.
So:
--- Given

Probability of winning would be:



The question illustrates binomial probability and will be solved using the following binomial expansion;

So:
Solving (a): Winning at least 1
We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4
So, we have:

Substitute value for W and L


<em>Hence, the probability of her winning at least one is 0.7599</em>
Solving (a): Wining exactly 2
We look at the above and we list out the terms where the powers of W is exactly 2
So, we have:

Substitute value for W and L


<em>Hence, the probability of her winning exactly two is 0.2646</em>