1/16........................................
Answer:
In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .
Answer:
The weights are 1 kg, 3kg, 9kg and 27kg.
Explanation:
The weights are 1 kg, 3kg, 9kg and 27kg.
1+3+9+27= 40
27+9+3= 39
27+9+3-1=38
27+9+1=37
27+9=36
27+9-1=35
27+9+1-3=34
27+9-3=33
27+9-3-1=32
27+3+1=31
27+3=30
27+3-1=29
27+1=28
27
27-1=26
27+1-3=25
27-3=24
27-3-1=23
27+3+1-9=22
27+3-9=21
27+3-9-1=20
Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.
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:
force is that push or pull of the body that change or tends to change the state of rest or uniform motion in a straight line.
Explanation:
hope it is helpful to u
