Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion


h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.
When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.
The part of the light that continues on into the new medium is <em>transmitted</em>
light. Its forward progress at any point in its journey is <em>transmission</em>.
Its direction usually changes as it crosses the boundary. The bending is <em>
refraction</em>.
The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light. The process of bouncing is <em>reflection</em>.
Northward
1810-1050=760
Accelerate 760 towards northward
Answer:
The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².
Explanation:
Given that,
Speed = 11.7 m/s
Coefficients of static friction = 0.48
Coefficients of kinetic friction = 0.34
Angle = 40.0°
(a). When the toboggan moves up hill, then
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula


(b). When the toboggan moves up hill, then
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula


Hence, The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².