Answer:
it can melt orcan put them past their boiling point
Explanation:
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
It's kind of funny really. The guy who invented the idea in the first place had actually been made fun of because the thought of that happening was improbable in their mind. Also, Wegener ( the guy who invented the idea) was not even a geologist but a meteorologist. But later it had been backed up due to earthquakes being record consecutively in one area, and the graphs turning out differently.
Answer:
1.6s
Explanation:
Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,
To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.
Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh
Where h = 5.2sin25
h = 2.2 m
P.E = 1.2 × 9.8 × 2.2
P.E = 25.84 j
At the bottom, K.E = P.E
1/2mv^2 = 25.84
Substitutes mass into the formula
1.2 × V^2 = 51.69
V^2 = 51.69/1.2
V^2 = 43.07
V = 6.56 m/s
Using the third equation of motion
V^2 = U^2 + 2as
Since the object started from rest,
U = 0
6.56^2 = 2 × a × 5.2
43.07 = 10.4a
a = 43.07/10.4
a = 4.14 m/s^2
Using the first equation of motion,
V = U + at
Where U = 0
6.56 = 4.14t
t = 6.56/4.14
t = 1.58s
Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s