Answer:
The three types of thermal expansions are Linear expansion , Superficial expansion and Cubical expansion
Answer:
If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds
Explanation:
First, let's put 22 km/h in m/s:

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
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 (→)
True. Think of a magnet and how they only connect to the opposite charges.