Answer:
σ = 0.255*10^-3 C/m²
Explanation:
The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.
At equilibrium ,
Upward force = downward force
Tcosθ = mg ----(1)
Assuming that the Forward force = backward force, then
Tsinθ = σq/ε₀
[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)
Dividing equation (2) by (1)
Tsinθ/Tcosθ = qσ/ε₀mg
⇒Tanθ = qσ/ε₀mg
σ = ε₀mg tanθ/q
Now substituting the values of
σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8
σ = (8.85*10^-12 * 0.5774) / 2*10^-8
σ = 5.11*10^-12 / 2*10^-8
σ = 0.255*10^-3 C/m²
Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2
Answer:
a. The temperature of the copper changed more than the temperature of the water.
Explanation:
Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:
(1)
But Q is related with mass (m), specific heat (c) and changes in temperature (
)in the next way:
(2)
Using (2) on (1):



Because we have an equality and 0.385 < 4.186 then
to conserve the equality
<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.