It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
Given :
An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.
To Find :
the image height (in cm).
Solution :
By lens formula :
Here, u = - 100 cm
f = 150 cm
Now, magnification is given by :
Therefore, the image height is 3 m or 300 cm.
Answer:
Chief Hopper
Explanation:
Mike travels at a constant speed of 3.1 m/s. To find how long it takes him to reach the school, we need to find the distance he travels. We can do this using Pythagorean theorem.
a² + b² = c²
(1000 m)² + (900 m)² = c²
c ≈ 1345 m
So the time is:
v = d / t
3.1 m/s = 1345 m / t
t ≈ 434 s
Next, Chief Hopper travels a total distance of 1900 m, starting at rest and accelerating at 0.028 m/s². So we can use constant acceleration equation to find the time.
d = v₀ t + ½ at²
1900 m = (0 m/s) t + ½ (0.028 m/s²) t²
t ≈ 368 s
So Chief Hopper reaches the school first, approximately 66 seconds before Mike does.
Answer: Surface water is replaced by the <u>water</u> cycle.
Explanation: The water cycle is a cycle that describes the movement of water.