Answer:
108 extended days
Explanation:
Regular school hours a day = 6 hr
No. of school days to make up by extending the regular hours = 3 days
Amount of time added to the regular hours of school = 10 min
No. of extended school days to make up the 3 school days by following the above mentioned criteria be x.
Time of school hours in 3 days =
are required to make up 3 days of school having 6 hours of regular timing with 10 minutes of add-on time each day.
Answer:
Magnitude of displacement = 2.07 km
Magnitude of average velocity = 1.17 kmph
Explanation:
Let east represent positive x axis and north represent positive y axis.
A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.
1.93 km due wast
s ₁ = 1.93 i km
1.03 km due south
s₂ = -1.03 j km
3.84 km in a direction 52.8 ° north of west
s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km
Total displacement
s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j
Magnitude of displacement,
Time taken = 1.771 hour
Magnitude of average velocity,
Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
