Using lens equation;
1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)
Substituting;
1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm
Therefore, the object should be place 99.23 cm from the lens.
I do not recall the answer to this question
Answer:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
We must remember that the total net force equation at
constant velocity is:
<span>F – Ff = 0</span>
of
F - µN = 0
Using Newton's 2nd Law of Motion:<span>
F = m a
<span>Where,
F = net force acting on the body
m = mass of the body
a = acceleration of the body
Since the cart is moving at a constant velocity, then
acceleration is zero, hence the working equation simplifies to
F = net Force = 0
Therefore,
F - µN = 0
where
µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N
Therefore,
F - 0.20(12) = 0
<span>
F = 2.4 N </span></span></span>
Answer:
The average power the woman exerts is 0.5 kW
Explanation:
We note that power, P = The rate at which work is done = Work/Time
Work = Energy
The total work done is the potential energy gained which is the energy due to vertical displacement
Given that the vertical displacement = 5.0 m, we have
Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h
m = 51 kg
g = Constant = 9.81 m/s²
h = 5.0 m
Also, time, t = 5.0 s
Total work done = 51 kg × 9.81 m/s²× 5 m = 2501.55 kg·m²/s² = 2501.55 J
P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.
