Answer:
Negative
Explanation:
Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).
Complete Question:
A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?
Answer:
Area = 0.00025 m²
Explanation:
Given the following data;
Pressure = 400N/m²
Force = 0.1N
To find the area of the coin;
Pressure = Force/area
Area = Force/pressure
Substituting into the equation, we have;
Area = 0.1/400
Area = 0.00025 m²
Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
Using the equation v(average)=x traveled/time
v = 100/2.5
You get 40 kilometers per hour
Hope this helped!
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
