The same voltage will appear across all resistors in parallel.
Answer:
78.4 m
Explanation:
Using newton's equation of motion,
S = ut + 1/2gt²......................... Equation 1
Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.
Note: Taking upward to be negative, and down ward positive
Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²
Substitute into equation 1
S = 49(2) - 1/2(9.8)(2)²
S = 98 - 19.6
S = 78.4 m
Hence the height of the ball two seconds later = 78.4 m
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = <em>360 km-m/s </em>
The direction of the change is <em>up /\ </em>.
Answer: The drag force goes up by a factor of 4
Explanation:
The <u>Drag Force</u> equation is:
(1)
Where:
is the Drag Force
is the Drag coefficient, which depends on the material
is the density of the fluid where the bicycle is moving (<u>air in this case)
</u>
is the transversal area of the body or object
the bicycle's velocity
Now, if we assume
,
and
do not change, we can rewrite (1) as:
(2)
Where
groups all these coefficients.
So, if we have a new velocity
, which is the double of the former velocity:
(3)
Equation (2) is written as:
(4)
Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>