You suspect that a power supply is faulty, but you use a power supply tester to measure its voltage output and find it to be acceptable. Why is it still possible that the power supply may be faulty?


Answer 1

While a power supply tester can be a useful tool for quickly checking voltage output, it might not reveal all the potential issues a faulty power supply can cause.

Even if a power supply tester shows that the voltage output of a power supply is within acceptable limits, it's still possible that the power supply may be faulty. Here's why:

1. Voltage Under Load: A power supply tester might only measure the voltage output under no load or very light load conditions.

A faulty power supply might provide the correct voltage at low loads but fail to deliver stable voltage under high loads, which could lead to system instability or crashes.

2. Voltage Ripple and Noise: Power supplies are expected to provide a stable and clean output voltage.

3. Short Circuits or Overloads: A power supply tester typically doesn't simulate the behavior of a real system.

4. Intermittent Issues: Faulty power supplies can exhibit intermittent issues. The power supply might work fine during the testing but fail when subjected to extended periods of operation or specific conditions.

5. Quality of Components: A power supply tester might not assess the quality of individual components within the power supply.

6. Compatibility Issues: Some power supplies might not be fully compatible with certain computer hardware. Even if the voltage seems fine, compatibility issues can still cause problems.

Learn more about Short Circuit here:


Related Questions

Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.
When an external magnetic field is applied, what happens to the protons in a sample?A) All protons align with the field.B) All protons align opposite to the field.C) Some protons align with the field and some align opposite to it.D) All protons assume a random orientation.
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2dB) d/√2C) d/4D) 2dE) d/2
Name the four forces in physics?​
What can you infer from the fact that metals are good conductors of electricity?

A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 1050 N/m and the mass of the rocket is 50 g, how high will the rocket go? You may neglect the effects of air resistance.



Rocket will go to a height of 8.678 m


Mass of the rocket m = 50 gram = 0.05 kg

Spring constant k = 1050 N /m

Spring is stretched to 9 cm

So x = 0.09 m

Work done in stretching the spring

E=(1)/(2)kx^2=(1)/(2)* 1050* 0.09^2=4.2525J

From energy conservation this energy will convert into potential energy

Potential energy is equal to E=mgh, here m is mass, g is acceleration due to gravity and h is height

So 0.05* 9.8* h=4.2525


So rocket will go to a height of 8.678 m


8.68 m


compression in spring, x = 9 cm = 0.09 m

Spring constant, K = 1050 N/m

mass of rocket, m = 50 g = 0.05 kg

Let it go upto height h.

Use conservation of energy

Potential energy stored in spring = potential energy of the rocket


0.5 x 1050 x 0.09 x 0.09 = 0.05 x 9.8 x h

h = 8.68 m

Thus, the rocket will go upto height 8.68 m.

In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.



L > 0.08944 m or L > 8.9 cm



- Flux intercepted by antenna Ф = 0.04 N.m^2 / C

- The uniform electric field E = 5.0 N/C


- What is the minimum side length of the antenna L ?


- We can apply Gauss Law on the antenna surface as follows:

                             Ф = \int\limits^S {E} \, dA

- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,

                             Ф = E.(L)^2

                             L = sqrt (Ф / E)

                             L > sqrt (0.04 / 5.0)

                             L > 0.08944 m

Final answer:

The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.


The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.

We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.

Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.

That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².

Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).

Learn more about Electric Flux here:


Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?


Hi, thank you for posting your question here at Brainly.

To solve this problem, we use Coulomb's Law:

F = kQ1Q2/d^2, where k = 9x10^9

Q1 = 3.5 uC
Q2 = -3.5 uC
Q3 = 4.0 uC

But first, we find the distance between Q1 and Q3 and between Q2 and Q3.

d between Q1 and Q2:
d = sqrt[(0-0.4)^2+(0.3-0)^2]
d = 0.5 m

d between Q1 and Q3:
d = sqrt[(0-0.4)^2+(-0.3-0)^2]
d = 0.5 m

Through force balance, F between Q2 and Q3 - F between Q1 and Q3:

F_(net) = ((9x 10^(9))(-3.5)(4) )/( 0.5^(2) ) -((9x 10^(9))(3.5)(4) )/( 0.5^(2) )=-1.008* 10^(12)

Thus, the net force is -1 x 10^-12 C

Final answer:

The total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC is zero. This is because the forces due to each of these charges on the third charge are equal in magnitude but opposite in direction, hence they cancel each other completely.


The question asks for the magnitude and direction of the total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC. This is related to Coulomb's Law, which describes the force between charged objects. Specifically, Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1*q2) and inversely proportional to the square of the distance (r) between them. It also depends on the permittivity of free space (ε₀).

First, you would determine the force between each of the point charges and the third charge separately, and then superpose these forces to find the total force. The force in each case can be calculated using the equation F = k*|q1*q2|/r², where k is Coulomb's constant (8.99 * 10^9 N.m²/C²). You would need to make sure you take into account the signs of the charges when deciding the directions of the forces and when superposing the separate forces.

Assume upwards to be the positive direction. The 3.5 uC charge forces and -3.5 uC charge forces on the 4 uC charge would be opposite in direction (one downwards and one upwards) and identical in magnitude. Therefore, they will cancel each other out, and hence, the total electric force on the third charge (4 uC) will be zero.

Learn more about Total Electric Force here:


Average wavelength of radio waves​


The average wavelength of radio waves ​ranges from roughly two millimeters to more than 150 kilometers. The wavelengths of radio waves are the longest in the electromagnetic spectrum

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

They also have the lowest frequencies, ranging from around 4,000 cycles per second, or 3 kilohertz, to roughly 280 billion hertz, or 280 gigahertz.

The wavelengths of radio waves are the longest in the electromagnetic spectrum, ranging from roughly two millimeters to more than 150 kilometers.

To learn more about wavelength from here, refer to the link given below;



Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.


Given Information:

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

Required Information:

Impulse = I = ?


Impulse = 11.77 kg.m/s


We know that impulse is equal to change in momentum

I = Δp

I = p₁ - p₂

I = mv₁ - mv₂

I = m(v₁ - v₂)

Where m is the mass of ball, v₂ is the final velocity of the ball, and v₁ is the initial velocity of the ball.

So first we need to find the initial and final velocities of the ball

The relation between initial potential energy and final kinetic energy before the collision is given by

PE₁ = KE₂

mgh₁ = ½mv₂²

gh₁ = ½v₂²

v₂² = 2gh₁

v₂ = √2gh₁

v₂ = √2*9.8*19

v₂ = 19.3 m/s

The relation between initial kinetic energy and final potential energy after the collision is given by

KE₁ = PE₂

½mv₁² = mgh₂

½v₁² = gh₂

v₁² = 2gh₂

v₁ = √2gh₂

v₁ =√2*9.8*15

v₁ = 17.15 m/s

Finally, we can now find the magnitude of the impulse delivered to the ball by the floor.

I = 0.320(17.5 - (-19.3))

I = 11.77 kg.m/s


Imp = 11.666\,(kg\cdot m)/(s)


Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (19\,m) = (1)/(2)\cdot (0.320\,kg)\cdot v_(A)^(2)

v_(A) \approx 19.304\,(m)/(s)

After collision:

(1)/(2)\cdot (0.320\,kg)\cdot v_(B)^(2) = (0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (15\,m)

v_(B) \approx 17.153\,(m)/(s)

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,(m)/(s) )-(-19.304\,(m)/(s) )]

Imp = 11.666\,(kg\cdot m)/(s)

A young man walks daily through a gridded city section to visit his girlfriend, who lives m blocks East and nblocks North of where the young man resides. Because the young man is anxious to see his girlfriend, his route to her never doubles back—he always approaches her location. In terms of m and n, how many different routes are there for the young man to take?



The man ate eggs.


He should brush his teeth before seeing his girlfriend.