They eat rice, chicken, beef, cereal they eat what we eat in a daily basis I guess
Explanation:
Given
Velocity v = 23.0m/s
Distance S = 3.45m
Required
Time it will take the skier to reach the ground;
Using the equation of motion;
S = ut + 1/2gt²
3.45 = 23t + 1/2(9.8)t²
3.45 = 23t + 4.9t²
4.9t²+23t-3.45 = 0
Factorize;
t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)
t = -23 ±√529+67.62/9.8
t = -23±√596.62/9.8
t = -23±24.43/9.8
t = 1.43/9.8
t = 0.146 secs
Hence take the skier 0.146 secs to reach the ground.
b) Horizontal distance covered is the range;
Range = U√2H/g
Range = 23√2(3.45)/9.8
Range = 23√6.9/9.8
Range = 23√0.7041
Range = 23(0.8391)
Range = 19.29m
Hence the horizontal distance travelled in air is 19.29m
Answer:
4117.65 N
Explanation:
Speed of ball, u = 126 km/h = 35 m/s
Mass of ball, m = 160 g = 0.16 kg
Time interval, t = 1.36 ms = 0.00136 s
We can calculate the force as a measure of the momentum of the ball:
F = P/t
Momentum, P, is given as:
P = mv
Therefore:
F = (mv) / t
F = (0.16 * 35) / (0.00136)
F = 4117.65 N
The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.
Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C