Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :




t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

Here, a = -g


h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
The shape is connected in parallel so;
5.1) Ans;

5.2) Ans;

I hope I helped you^_^
Answer:
22.11 m / s
Explanation:
The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )
initial velocity of falcon = 28 cos 35 i - 28 sin 35 j
( falcon was flying in south east direction making 35 degree from the east )
momentum = .9 ( 28 cos 35 i - 28 sin 35 j )
= 20.64 i - 14.45 j
initial velocity of pigeon
= 7 i
initial momentum = .325 x 7i
= 2.275 i
If final velocity of composite mass of falcon and pigeon be V
Applying law of conservation of momentum
( .9 + .325) V = 20.64 i - 14.45 j +2.275 i
V = ( 22.915 i - 14.45 j ) / 1.225
= 18.70 i - 11.8 j
magnitude of V
= √ [ (18.7 )² + ( 11.8 )²]
= 22.11 m / s
Answer:
600km/h as u are on a platform moving at the speed of 600 km/h where u are moving in relativity to the plane it's self.
Answer:
$ 1.1
Explanation:
From the question given above, the following data were obtained:
Cost per kWh = $ 0.1
Current (I) = 10 A
Voltage (V) = 220 V
Time (t) = 5 h
Cost of operation =?
Next, we shall determine the power the electric oven. This can be obtained as follow:
Current (I) = 10 A
Voltage = 220 V
Power (P) =?
P = IV
P = 10 × 220
P = 2200 W
Next, we shall convert 2200 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
2200 W = 2200 W × 1 KW / 1000 W
2200 W = 2.2 KW
Thus, 2200 W is equivalent to 2.2 KW.
Next, we shall determine the energy consumed by the electric oven. This can be obtained as follow:
Power (P) = 2.2 KW
Time (t) = 5 h
Energy (E) =?
E = Pt
E = 2.2 × 5
E = 11 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost $ 0.1
Therefore,
11 KWh will cost = 11 × 0.1
11 KWh will cost = $ 1.1
Therefore, the cost of operating the electric oven is $ 1.1