Answer:
Option D.
Explanation:
Methods of charging:
1. By friction: when two materials of different electro-negativities rub against each, electrons transfer from one material to another.
2. By induction: When a charged body is brought near an uncharged body, the later gets polarized.
3. By conduction: when electrons from one body to another uncharged while contact, it is known as charging by conduction.
Here, a flower is analogous to an electron. A box of candy is positive charge. A bouquet of flowers is like a charged body which is brought near an uncharged body. It induces charge on ti it by causing it to transfer electron to another.
Thus, option D is correct.
With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.
Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2
Then the equation for velocity, V is
V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t
Vo is the initial velocity, which you can find using V = 54km/h and t = -2
Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h
Then, the equation is: V = 50 km/h + 2t
Valid for constant acceleration.
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
Answer:
Explanation:
7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s
7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m
or
h = (45sin14.5)² / (2(9.81)) = 6.47 m
which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.
