Answer: 355.66 m/s
Explanation:
Since sound is a mechanical wave, its speed varies with the medium and its properties, such as its temperature.
So, if we want to calculate the speed of sound in air at we can use the following equation that relates with the temperature:
(1)
Where:
is the Heat capacity ratio for air
is the Universal Gas constant
is the air temperature in Kelvin
is the air molar mass
(2)
Finally:
This is the speed of sound at 313 K
Explanation:
For each point:
PE = mgh
KE = ½ mv²
ME = PE + KE
Since energy is conserved, ME will be constant.
I'll be using g = 10 m/s².
A. PE = (100 kg) (10 m/s²) (102 m) = 102,000 J
KE = ½ (100 kg) (0 m/s)² = 0 J
ME = 102,000 J
B. KE = ½ mv²
18,000 J = ½ (100 kg) v²
v = 60 m/s
ME = 102,000 J
PE = ME − KE = 84,000 J
(100 kg) (10 m/s²) h = 84,000 J
h = 84 m
C. KE = ½ (100 kg) (29 m/s)² = 42,050 J
ME = 102,000 J
PE = ME − KE = 59950 J
(100 kg) (10 m/s²) h = 59950 J
h = 59.95 m
d. PE = (100 kg) (10 m/s²) (60 m) = 60,000 J
ME = 102,000 J
KE = ME − PE = 42,000 J
½ (100 kg) v² = 42,000 J
v = 29.0 m/s
E. PE = (100 kg) (10 m/s²) (0 m) = 0 J
ME = 102,000 J
KE = ME − PE = 102,000 J
½ (100 kg) v² = 102,000 J
v = 45.2 m/s
Current is charge over time
C = Q/t
C = 96 ÷ 12
C = 8
8 Amperes
Answer:
yi = Initial height of the helicopter
yf = final height of the helicopter
vyi = component of the initial vertical velocity of the helicopter
g = gravity constant (9.8m/s^2)
yf = yi + vyideltat - 1/2gt^2
0m = 1000m + (15m/2)deltat - 1/2(9.8m/s^2)t^2
-1000m = (15m/s)t - (-4.9m/s^2)t^2
Use the quadratic formula
4.8t^2 - 15t - 1000 = 0
t1 = 15.75s and t2 = -12.65
t2 is rejected, time can't be negative
Thus, it takes 15.75s before the package strikes the ground.