Answer:
Explanation:
Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.
Givens
=====
V
= 4.00 L
T
= 273oK We're assuming the temperature does not change, just the
pressure.
n
= 0.864 moles
R
= 8.314 joules / mole * oK
P
= ?????
Formula
======
PV
= n*R*T
P
= n*R*T/V
P
= 0.864 * 8.314 * 273 / 4
P
= 490 kpa
You
have to add 1.6 – 0.864 = 0.736 moles of gas.
We
have to assume that the temperature and pressure remain the same when
we add the 0.736 moles of gas. We are now looking for the volume.
PV
= n*R*T
<span>
V
= 0.736 * 8.314 * 273 / 490</span>
V
= 3.41 L Remember this is at about 4 atmospheres so we have to
convert to Standard Pressure.
Total
Volume = 3.41 + 4.00 = 4.41
V1
* P1 = V2 * P2
P1
= 490 kPa
P2
= 101 kPa
V1
= 7.41 L
V2
= ????
<span>
<span>
7.41*
490 = V2 * 101
V2
= 7.41 * 490 / 101
V2
= 35.94 L
</span>
</span>
<span>You
had 4 L now you need 31.94 more.</span>
Answer:
F = 6.27 x 10 ¹⁹ N
Explanation:
Given
m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol
q₁ = % * [m * N * A * e / M ]
q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₁ = 3.54 x 10⁶ C
q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₂ = 1.773 x 10⁶ C
Now to determine the electrostatic force con use the equation
F = K * q₁ * q₂ / d²
K = 8.99 x 10 ⁹
F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²
F = 6.27 x 10 ¹⁹ N
Answer:
18%
Explanation:
There are two equal and opposite forces on a floating object: weight and buoyancy.
W = B
The weight of an object is its mass times gravity: W = mg
Buoyancy is the weight of the displaced fluid: W = mf g
Plugging in:
mg = mf g
m = mf
Mass is density times volume:
ρV = ρf Vf
Solving for the ratio of Vf / V:
Vf / V = ρ / ρf
Given that ρ = 0.82 g/mL and ρf = 1.00 g/mL:
Vf / V = 0.82
That means 82% of the object's volume (and therefore, 82% of its mass, assuming uniform density) is submerged. Which means that 18% is above the water line.
