Distance= Time×Speed
= 1800×1.5
= 2700 m
I am not sure it's right. the question itself is confusing.
Answer:
Explanation:
Given
fly is at height of
mouse is at a distance of 30 ft from bush
if is the angle of elevation then
using trigonometric relation
Answer:
Approximately at below the horizon.
- Horizontal component of velocity: .
- Vertical component of velocity: (downwards.)
(Assumption: air resistance on the ball is negligible; .)
Explanation:
Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at until the ball reaches the ground.
On the other hand, the vertical component of the ball would increase (downwards) at a rate of (where is the acceleration due to gravity.) In , the vertical component of the velocity of this ball would have increased by .
However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was . Hence, after the ball rolled off the table, the vertical component of the velocity of this ball would be .
Calculate the magnitude of the velocity of this ball. Let and and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be .
At after the ball rolled off the table, while . Calculate the magnitude of the velocity of the ball at this moment:
.
Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let denote that angle.
.
For the vector representing the velocity of this ball:
.
Calculate the size of this angle:
.
Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.
Answer:
M = 222 fringes
Explanation:
given
λ = 559 n m = 559 × 10⁻⁹ m
radius = 0.026 mm = 0.026 ×10⁻³ m
length of the glass plate = 22.1 ×10⁻² m
using relation
= 221.79
= 221 (approx.)
hence no of bright fringe
M = m + 1
= 221 +1
M = 222 fringes