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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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1-Calcular la velocidad de un móvil que puede recorrer 108 Km en 3h, utilizando las ecuaciones aprendida.
Ede4ka [16]

Answer:

L*W *H

Explanation:

5 0
3 years ago
135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the ident
telo118 [61]

Answer:

The unknown substance is Aluminum.

Explanation:

We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:

Initial temperature (T₁) = 25 ⁰C

Final temperature (T₂) = 100 ⁰C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 25

ΔT = 75 ⁰C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Change in temperature (ΔT) = 75 ⁰C

Mass of the substance (M) = 135 g

Heat (Q) gained = 9133 J

Specific heat capacity (C) of substance =?

Q = MCΔT

9133 = 135 × C × 75

9133 = 10125 × C

Divide both side by 10125

C = 9133 / 10125

C = 0.902 J/gºC

Thus, the specific heat capacity of substance is 0.902 J/gºC

Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.

7 0
3 years ago
What is the average horizontal component of force exerted on his feet by the ground during acceleration?
Arte-miy333 [17]
Well you have to think of it like electricity go through your answer closes to that and figure it out
7 0
3 years ago
From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft
zhannawk [14.2K]

Answer:

W=\frac{773}{4.45}=173.76 l b f

Explanation:

W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}

The law of gravitation

G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)

Universal gravitational constant [S.I. units]

m_{e}=5.976\left(10^{24}\right) k g

Mass of Earth [S.I. units]

m=89 kg

Mass of a man in a spacecraft [S.I. units]

R=6371 \mathrm{~km}

Earth radius [km]

Distance between man and the earth's surface

h=261 \mathrm{~km} \quad[\mathrm{~km}]

ESULT W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}

W=\frac{773}{4.45}=173.76 l b f

4 0
2 years ago
Explain what happens to the pitch of a cell phone ring when the amplitude of a sound wave increases. Justify your reasoning usin
dalvyx [7]
<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases. 

Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>
7 0
2 years ago
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