<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Here the ball travels 3 m less distance in fifth second compared to third second.
That is
s₃ = s₅ + 3
Now we have
Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²
s₃ = u - 2.5 g
Also
Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²
s₅ = u - 4.5 g
That is
u - 2.5 g = u - 4.5 g + 3
2 g = 3
g = 1.5 m/s²
Acceleration due to gravity in moon = 1.5 m/s²
Answer:
Explanation:
Energy of an inductor = 1/2 L i²
L is inductance , i is current .
= 1/2 x 12 x .3²
= .54 J
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
Answer:
The value is 
or 21.45%
Explanation:
From the question we are told that
The first reservoir is at steam point 
The second reservoir is at room temperature 
Generally the maximum theoretical efficiency of a Carnot engine is mathematically evaluated as
=> 
=>
Answer:
It looks like its moving north.
Explanation:
