Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
Answer:
The acceleration is a = 2.75 [m/s^2]
Explanation:
In order to solve this problem we must use kinematics equations.
where:
Vf = final velocity = 13 [m/s]
Vi = initial velocity = 2 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Now replacing:
13 = 2 + (4*a)
(13 - 2) = 4*a
a = 2.75 [m/s^2]
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
Answer:
r = 4.44 m
Explanation:
For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid
B = ρ g V
Now let's use Newton's equilibrium relationship
B - W = 0
B = W
The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)
σ = W / A
W = σ A
The area of a sphere is
A = 4π r²
W = W₁ + σ 4π r²
The volume of a sphere is
V = 4/3 π r³
Let's replace
ρ g 4/3 π r³ = W₁ + σ 4π r²
If we use the ideal gas equation
P V = n RT
P = ρ RT
ρ = P / RT
P / RT g 4/3 π r³ - σ 4 π r² = W₁
r² 4π (P/3RT r - σ) = W₁
Let's replace the values
r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000
r² (11.81 r -0.060) = 13000 / 4pi
r² (11.81 r - 0.060) = 1034.51
As the independent term is very small we can despise it, to find the solution
r = 4.44 m
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
