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2 years ago

9

Assume your computer has a power rating of 50.0 W, and you use your computer for 7.0 hours a day for a normal school day. If the

electric company charges $0.10 per kWh, how much does it cost you to use your computer each school day?
35$

$0.035

$0.005

$0.35

1 answer:

Vlad [161]2 years ago

6 0

Answer is in the link

HTTPS:/www.word doc.com

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Objects that transmit some light and reflect/absorb the rest of the light, like stained glass, are

11111nata11111 [884]

You are correct it is translucent because opaque let’s no light through but transparent lets almost all light through

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3 years ago

How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?

slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

<em>applied force, F = 600 N</em>
<em>mass of the truck, m = 1,500 kg</em>
<em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

$F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s$

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

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2 years ago

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
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2 years ago

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The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J$

As the energy in the system is conserved we have

$\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s$

The speed of the 8 kg block just before collision is 3.258 m/s

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3 years ago

30 points + brainliest

pishuonlain [190]

Answer:B is the answer

4 0

2 years ago

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