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3 years ago

8

One summer i installed some decorative low voltage lighting (15 volt) around my patio. the lights are wired in parallel. the man

ufacturer says that the power supply can safely deliver 8 a of current. how many 18 watt lights can i attach to a single power supply?

1 answer:

hammer [34]3 years ago

5 0

First, we determine the power that the assembly can supply by multiplying the voltage by the current. That is,

Power = voltage x current

Substituting the known values,

P = (15 V)(8 a)
P = 120 watts

To determine the answer to this item, we divide the calculate power of each,

n = 120 watts / 18 watt/light
n = 6.67 lights

The answer to this item is 6.

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3 years ago

How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and th

alexandr1967 [171]

Answer:

t = 4.58 s

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3 years ago

In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

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Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

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Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

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Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .

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RUDIKE [14]

(a)
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To find to how many megatons the meteorite energy loss $\Delta E$
corresponds, we can set the following proportion
$1 MT: 4.2 \cdot 10^{15}J=x: \Delta E$
And so we find
$x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT$
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of $\Delta E=0.162 MT=162 kT$ . So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
$1:13 kT=x:162 kT$
And so we find
$x= \frac{162 kT}{13 kT}=12.46$
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.

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4 years ago