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
Microwaves are transmitted by Radio Waves.
Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>
T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>
T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
B. By vibrations in wires or strings
