Answer:
It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.
Explanation:
Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by

Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get

Solving the above equation we get
we N= 2.313≅ 2
Answer:
t = 0.714 s and x = 5.0 m
Explanation:
This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed
vₓ = 7.0 m / s
Let's find the time it takes to get to the river
y = y₀ + v_{oy} t - ½ g t²
the initial vertical speed is zero and when it reaches the river its height is zero
0 = y₀ + 0 - ½ g t²
t =
t = ra 2 2.5 / 9.8
t = 0.714 s
the distance traveled is
x = vₓ t
x = 7.0 0.714
x = 5.0 m
Answer:
B) Pressure on the scale, not registered as weight.
Explanation:
This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point
Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.