Answer:
House A requires heat at a slightest faster rate than B
Explanation:
House A requires heat at a slightest faster rate than B due to the slight high temperature the furnace A is.
Answer:
<h2>
The magnitude of force F is 18N</h2>
Explanation:
The magnitude of the force in the set up can be solved for using the principle of moment. According to the principle, the sum of clockwise moment is equal to the sum of anticlockwise moments.
Moment = Force * perpendicular distance
Clockwise moments;
The force that acts clockwise is the unknown Force F and 4N force. If the beam rests on a pivot 60 cm from end X and a Force F acts on the beam 80 cm from end X, the perpendicular distance of the force F from the pivot is 80-60 = 20cm and the perpendicular distance of the 4N force from the pivot is 60-50 = 10cm
Moment of force F about the pivot = F * 20
Moment of 4N force about the pivot = 4*10 = 40Nm
Sum of clockwise moment = 40+20F...(1)
Anticlockwise moment;
The 8N will act anticlockwisely about the pivot.
The distance between the 8N force and the pivot is 60-10 = 50cm
Moment of the 8N force = 8*50
=400Nm...(1)
Equating 1 and 2 we have;
40+20F = 400
20F = 400-40
20F = 360
F = 18N
The magnitude of force F is 18N
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
Answer:
A) The free body diagrams for both the load of bricks and the counterweight are attached.
B) a = 2.96 m/s²
Explanation:
A)
The free body diagrams for both the load of bricks and the counterweight are attached.
B)
The acceleration of upward acceleration of the load of bricks is given by the following formula:
a = g(m₁ - m₂)/(m₁ + m₂)
where,
a = upward acceleration of load of bricks = ?
g = 9.8 m/s²
m₁ = heavier mass = mass of counterweight = 28 kg
m₂ = lighter mass = mass of load of bricks = 15 kg
Therefore, using these values in equation, we get:
a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)
<u>a = 2.96 m/s²</u>
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m
