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3 years ago

12

How much energy is dissipated as heat during a two-minute time interval by a 1.5- kΩ resistor which has a constant 20- V potenti

al difference across its leads?

1 answer:

lord [1]3 years ago

7 0

Answer:

32 J

Explanation:

Power is V²/R = 20²/1500 = 4/15 . . . watts

In 120 seconds, the energy is ...

(4/15 J/s)×(120 s) = 32 J

In a 2-minute period 32 joules are dissipated as heat.

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In which city did the wright brothers take off on their very first flight

Nezavi [6.7K]

<h2>Answer: Kitty Hawk, North Carolina </h2>

The Wright brothers, Wilbur and Orville, were pioneers of aviation, since they flew in a device heavier than air, which was inconceivable at that time.

Their first successful flight was on December 17th, 1903 in Kitty Hawk, North Carolina, which lasted only 12 seconds in which their plane (the Flyer I, with 341 kg, 6.4 m long and a wingspan of 12.3 m) traveled 37 m without touching the ground. This was achieved through the help of an external catapult that "threw" them into the air.

It should be noted that the Wright brothers only studied until high school, however, their passion for solving the problem of the human inability to fly, their perseverance and experience acquired over the years in their bicycle company, led them to reach that goal. An achievement that marked the beginning of the aviation era.

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3 years ago

A hydrogen atom has a diameter of about 10 nm. Express this diameter in meters. Express this diameter in millimeters. Express th

Fynjy0 [20]

So, the first question is: how many meters are 10 nm?

1nm =<span>0.000000001 m.

So 10 nanometers are </span><span>0.00000001 m!

Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!

now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
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3 years ago

A car with a mass of 2.0×10^3 kg is traveling at 15m/s .what is the momentum of the car ?

Maurinko [17]

Hello,

<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:

p=mv

Where,

p = momentum
m = mass
v = </span>velocity

The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:

p=2.0 x 10^3 *15 or 2000(15)

p=30000

Thus, the cars momentum is 30000 kg m/s

Faith xoxo

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3 years ago

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

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3 years ago

A length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

ycow [4]

The equivalent resistance of resistors connected in parallel is
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$
In our problem, we have 5 identical resistor of resistance $R$ , so their equivalent resistance is
$\frac{1}{R_{eq}} = 5 (\frac{1}{R} )$
The problem also says that the equivalent resistance is $R_{eq}= 17 \Omega$ , so we can find the resistance R of each piece of wire:
$R=5 R_{eq} = 5 \cdot 17 \Omega = 85 \Omega$

In the initial wire, it's like the 5 pieces are connected in series, so the equivalent resistance of the wire is just the sum of the resistances of the 5 pieces:
$R_{wire} = 5 R = 5 \cdot 85 \Omega = 425 \Omega$

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3 years ago