Answer: height = 3.98m
Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae
p = mgh
p = potential energy = 4.61kJ = 4610J
m = mass of watermelon = 118 kg
g = acceleration due gravity = 9.8 m/s²
4610 = 118 * 9.8 * h
h = 4610/ 118 * 9.8
h = 4610/ 1156.4
h = 3.98m
The correct answer would be "He brought one serving to his neighbor's house, and stored the other two servings in the refrigerator. Devon ate one more serving or spaghetti the following day."
Answer:
a) load in Newton is 96,138 b) 129.314mm
Explanation:
Stress = force/ area (cross sectional area of the bronze)
Force(load) = 294*10^6*327*10^-6 = 96138N
b) modulus e = stress/ strain
Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3
Strain = change in length/ original length = DL/ 129
Change in length DL = 129 * 2.34*10^ -3 = 0.31347
Maximum length = change in length + original length = 129.314mm
Answer
D.Diffraction
Explanation
Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.
The ways tends to curve behind the barrier. This is called diffraction of waves.
Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
