Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
Answer:
The velocity is 60 km/hr.
Explanation:
<h3><u>Given:</u></h3>
Displacement (d) = 480 km = 48000 m
Time (t) = 8 Hours = 480 minute
Velocity (v) = ?
Now,
Velocity = Displacement ÷ Time
v = d/t
v = 480/8
v = 60 km/hr
Thus, The velocity is 60 km/hr.
<u>-TheUnknownScientist 72</u>
So the result is not biased or affected in some way
Answer:
163.33 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 40 Kg
Height (h) = 25 m
Time (t) = 1 min
Power (P) =..?
Next, we shall determine the energy. This can be obtained as follow:
Mass (m) = 40 Kg
Height (h) = 25 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
E = mgh
E = 40 × 9.8 × 255
E = 9800 J
Finally, we shall determine the power. This can be obtained as illustrated below:
Time (t) = 1 min = 60 s
Energy (E) = 9800 J
Power (P) =?
P = E/t
P = 9800 / 60
P = 163.33 Watts
Thus, the power required is 163.33 Watts
Answer:
Explanation:
Total length of the wire is 29 m.
Let the length of one piece is d and of another piece is 29 - d.
Let d is used to make a square.
And 29 - d is used to make an equilateral triangle.
(a)
Area of square = d²
Area of equilateral triangle = √3(29 - d)²/4
Total area,
![A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}](https://tex.z-dn.net/?f=A%20%3D%20d%5E%7B2%7D%2B%5Cfrac%7B%5Csqrt3%7D%7B4%7D%5Cleft%20%28%2029-d%20%5Cright%20%29%5E%7B2%7D)
Differentiate both sides with respect to d.
![\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D2d-%20%5Cfrac%7B%5Csqrt3%7D%7B4%7D%5Ctimes%202%2829-d%29)
For maxima and minima, dA/dt = 0
d = 8.76 m
Differentiate again we get the
![\frac{d^{2}A}{dt^{2}}= + ve](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7DA%7D%7Bdt%5E%7B2%7D%7D%3D%20%2B%20ve)
(a) So, the area is maximum when the side of square is 29 m
(b) so, the area is minimum when the side of square is 8.76 m