Answer:
9.461 × 10^12 Km s
..............................
Answer:
The pressure of the air molecules inside the pen cap increases and the volume occupied by the air decreases such that the combined volume occupied by the pen cap and the air volume reduces while the combined mass of the pen cap and the air molecules remain the same
Given that density = The mass/Volume, we have that the density varies inversely as the volume, and as the volume reduces, the density increases
Upon squeezing, therefore, as the new combined density of the pen cap and the air molecules rises to more than the density of the water in the bottle, then, the pen cap air molecule is relatively more denser than the water, which will result in the pen cap sinking to the bottom of the bottle
Explanation:
Answer:
ΔV = 20.1 V
Explanation:
As the positive plates are connected to each other, the capacitors are connected in parallel, so the total system load is the sum of the charges on each capacitor.
Q = Q₁ + Q₂
The charge on each capacitor is
Q₁ = C₁ ΔV₁
Q₁ = 24 10⁻⁶ 25
Q₁ = 6.00 10⁻⁴ C
Q₂ = C₂ ΔV₂
Q₂ = 13 10⁻⁶ 11
Q₂ = 1.43 10⁻⁴ C
The total set charge is
Q = (6 + 1.43) 10⁻⁴
Q = 7.43 10⁻⁴ C
The equivalent capacitance is
C_eq = C₁ + C₂
C_eq = (24 + 13) 10⁻⁶
C_eq = 37 10⁻⁶ F
Let's use the relationship to find the voltage
Q = C_eq ΔV
ΔV = Q / C_eq
ΔV = 7.43 10⁻⁴ / 37 10⁻⁶
ΔV = 2.008 10¹
ΔV = 20.1 V
This voltage is constant in the combination so it is also the voltage in capacitor C1
Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R =
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as =
Taking the integral of both sides;we have:
R = 14.863 Ω
Since V = IR
I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
Answer:
216.31 (the work done by gravity is -216.31) positive for going up.
Explanation:
We look at this question first by getting the right equation for <em>work</em>.
Which should be... W = F x D.
From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.
What is the formula for the force of gravity on an object near the earth?
Right here --- = mg
m = the mass and...
g = the acceleration due to gravity which is <em>9.81 m/s2</em>
Before we plug things in though, we need to convert everything to SI units,
the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.
Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) = 216.31 J or 216.31 N m.