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Klio2033 [76]
3 years ago
14

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 acc

eptable. Why is it still possible that the power supply may be faulty?
Physics
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

Load

Explanation:

A normal power supply can deliver up to certain amount of power to a load. The output power can be calculated multiplying Voltage (V) x Current (A). It happens that after a certain period of time, the power source's main components begin to wear, thus losing its ability to deliver its nominal power. Normally, when no load its connected to the source, you will get the operating Voltage, but when the load demands power, the ability to deliver power to it may fail to reach nominal levels. When connected, there may be voltage drops (thus, less power output) causing malfunctions turning it into a non-operative power supply.

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Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

4 0
3 years ago
You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
What is the acceleration of a car that goes from 0 to 20m/s in 7 seconds?
nadezda [96]

Answer:

2m/s^2

Explanation:

Clculate the acceleration:

V = u +at

20m/s = 0 + a*10s

a = 20m//10s

a = 2m/s²

From the data given , it is not possible to calculate the displacement , because no direction of motion is given

But it is possible to calculate the distance travelled

Distance = ut + ½ *a*t²

distance = 0 + ½ * 2m/s * 10²s

distance = 100m

6 0
2 years ago
A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?
Rama09 [41]
The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J
4 0
3 years ago
Read 2 more answers
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

6 0
3 years ago
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