Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
How fast and efficient the energy is released.
Explanation:
Before burning the marshmallow energy is stored in it in the form of chemical bond energy or chemical potential energy. So upon burning this energy is released. So there will be a difference in energy release from a burned marshmallow and the one we eat straight from package.
Answer:
Sound waves. Anything that vibrates is producing sound; soundis simply a longitudinal wave passing through a medium via the vibration of particles in themedium. Consider a sound wavetraveling in air
The approximate lateral area of the prism is determined as 831 square inches.
<h3>
What is lateral area of the hexagonal prism?</h3>
The lateral area of the hexagonal prism is calculated as follows;
LA = PH
where;
- P is perimeter of the prism
- H is height
A = ¹/₂Pa
where;
- a is apothem = 10 inches
- A is base area = 346.41 in²
346.41 = ¹/₂(10)P
346.41 = 5P
P = 346.41/5
P = 69.282 inches
LA = PH
LA = 69.282 x 12
LA = 831.38 in²
Learn more about lateral area of prism here: brainly.com/question/296674
#SPJ4
Answer: you cant see sound waves but youcan defiently hear them . when the travle through difrent levels they depend on how loud the sound wave is if you hear a loud sound its called a loud sound wave
Explanation:
