Answer:
<h2>9.375Nm</h2>
Explanation:
The formula for calculating torque τ = Frsin∅ where;
F = applied force (in newton)
r = radius (in metres)
∅ = angle that the force made with the bar.
Given F= 25N, r = 0.75m and ∅ = 30°
torque on the bar τ = 25*0.75*sin30°
τ = 25*0.75*0.5
τ = 9.375Nm
The torque on the bar is 9.375Nm
Answer:
a) Focal length of the lens is 8 cm which is a convex lens
b) 6 cm
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Explanation:
u = Object distance = 4 cm
v = Image distance = -8 cm
f = Focal length
Lens Equation
a) Focal length of the lens is 8 cm which is a convex lens
Magnification
b) Height of image is 2×3 = 6 cm
Since magnification is positive the image upright
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Answer:
I1 = 0.772 A
Explanation:
<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts
<u>To find</u>: current I = ? A
<u>Solution: </u>
Ohm's law V= I R
⇒ I = V / R
In order to find R (total) we first find R (p) fro parallel combination. so
1 / R (p) = 1 / R1 + 1/ R2 ∴(P) stand for parallel
R (p) = R1R2 / ( R1 + R2)
R (p) = (5.0 × 9.0) / (5.0 + 9.0)
R (p) = 3.214 ohm
Now R (total) = R (p) + R3 (as R3 is connected in series)
R (total) = 3.214 ohm + 4.0 Ohm
R (total) = 7.214 ohm
now I (total) = 7.214 ohm / 6.0 Volts
I (total) = 1.202 A
This the total current supplied by 6 volts battery.
as voltage drop across R (p) = V = R (p) × I (total)
V (p) = 3.214 ohm × 1.202 A = 3.864 volts
Now current through 5 ohms resister is I1 = V (P) / R1
I1 = 3.864 volts / 5 ohm
I1 = 0.772 A
Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
Answer:
(b) 10 Wb
Explanation:
Given;
angle of inclination of magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
Magnetic flux is given as;
Φ = BACosθ
where;
B is the strength of magnetic field
A is the area of the plane
θ is the angle of inclination
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Now calculate the magnetic flux through a 2.0 m² portion of the same plane
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.
Option "b"
