Answer:
that technician A is right
Explanation:
The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.
From the above it is seen that technician A is right
Answer:
The answer is below
Explanation:
Given that:
mass (m) = 86 kg, distance (L) = 2.75 m, θ = 31°, force (F) = 595 N, initial velocity (
) = 2.4 m/s, g = acceleration due to gravity = 9.8 m/s²
The net work can be gotten from the equation:
From the work-energy theorem equation, we can get her speed at the top of the ramp (
)
Hence:
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
The temperature and the solubility of sugar at that temperature
Explanation:
The amount of substance which can be dissolved in the solvent depends on the temperature.
As the temperature increases, more substance can be dissolved.
A solution is saturated if any more of the solute cannot be dissolved in the solution at the given temperature
Hence we need to know the temperature and also the amount of substance which can be dissolved(solubility) at the same temperature
a) the statement given in option A is correct
b) molar mass has no correlation with the substance's solubility and hence option b is not correct
c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
