Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.
Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is
Square root of (19.6h) meters per second.
It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, 
After the collision, final momentum 
Using the conservation of momentum as :
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
Answer:
Explanation:
We know that the pressure can be calculated in the following way:
p = d·g·h
with d being the density of the water, g the gravitational acceleration and h the depth.
Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa
117,500 Pa = 1000kg/m³ · 9.8m/s² · h
Therefore h = 11,9 m
I’m thinking the Last one tbh
