A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speed will the sled have twice as much kinetic energy?
4.2 m/s
2 meters per second for 8 seconds
1.5 meters per second for 8 seconds
The average speed should be 1.8 or 1.7 but I think its 1.8
It's called gliding.
So, they will get up high and collect their balance and be nice and steady. Then, they can spread their wings out and glide. The wind will carry them from their.
How do we manage to stand without constantly walking? It's the same exact thing.
Answer:
Explanation:
Given that,
Diameter of pipe
d = 20cm
Then, radius =d/2 =20/2
r = 10cm =0.1m
The speed at the bottom is
Vi = 3m/s
Speed at the top Vf?
At the bottom the cube is at a height of 0m
Then, y1 = 0m
At the top the cube is at a height which is the same as the diameter of the pipe
y2 = 0.2m
Now, let us consider, the energy conservation equation , which is the sum of kinetic energy and gravitational potential energy, given by,
K2 + U2 = K1 + U1
½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1
Divide all through by m
½•Vf² + g•y2 = ½•Vi² + g•y1
Since y1 = 0
So we have,
½•Vf² + g•y2 = ½•Vi²
½•Vf² = ½•Vi² — g•y2
Multiply through by 2
Vf² = Vi² —2g•y2
Vf = √(Vi²—2g•y2)
g is a constant =9.81m/s2
Vf = √(3²—2×9.81×0.2)
Vf = √(9—0.981)
Vf = √8.019
Vf = 2.83m/s
The speed of the ice cubes at the top of the pipe is 2.83m/s
Using kinematics we can find that the take-off distance is 6163 ft
Given parameters
- The initial and final speed of the plane i = 0 and v = 140 mph
To find
The measurement system allows not to have problems when working in different units, in this case we reduce the speed units
v = 140 mile / h (5280 ft / mile) (1h / 3600 s) = 205.34 ft / s
The kinematics allows to find the relationships between the position, the speed and the acceleration of a body, in this case the movement is in one dimension.
v = v₀ + a t
where v and v₀ are the final and initial velocity, respectively, at acceleration and t the time
a =
a =
a = 3.42 ft / s²
Let's use the expression
v² = v₀² + 2 a x
Where v and v₀ are the final and initial velocity, respectively, at acceleration and x the distance traveled
x =
x =
x = 6163.8 ft
Let's reduce to miles
x = 6163.8 ft (1 mile / 5280 ft)
x = 1.17 mile
In conclusion using kinematics we can find that the take-off distance is 6163 ft
Learn more about kinematics here: