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xxMikexx [17]
3 years ago
12

Power dissipated is 40w in the 10 ohms resistor. What is the

Physics
1 answer:
sveticcg [70]3 years ago
3 0

tiodoododoodkdkdkdedd

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A satellite in orbit is not truly traveling through a vacuum.It is moving through very, very thin air. Does the resulting airfri
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PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.

However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.

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4 years ago
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Katena32 [7]
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4 0
4 years ago
Particle moves in a circle of radius 90m with a constant speed 25m/s. how many revolution does it make in 30sec​
Serggg [28]

Answer:

<em>n =1.33 revolutions</em>

Explanation:

<u>Uniform Circular Motion</u>

The angular speed can be calculated in two different ways:

\displaystyle \omega=\frac{v}{r}

Where:

v = tangential speed

r = radius of the circle described by the rotating object

Also:

\omega=2\pi f

Where:

f = frequency

Solving for f:

\displaystyle f=\frac{\omega}{2\pi}

Since the frequency is calculated when the number of revolutions n and the time t are known:

\displaystyle f=\frac{n}{t}

We can solve for n:

n=f.t

The particle moves in a circle of r=90 m with a speed v=25 m/s. Thus the angular speed is:

\displaystyle \omega=\frac{25}{90}

\displaystyle \omega=0.278\ rad/s

Now we calculate f:

\displaystyle f=\frac{0.278}{2\pi}

f=0.04421\ Hz

Calculating the number of revolutions:

n = 0.04421*30

n =1.33 revolutions

8 0
3 years ago
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Pavlova-9 [17]
C might be the answer
3 0
3 years ago
Read 2 more answers
How to solve 2.14 using calculus
trasher [3.6K]

Answer:

The acceleration is 6.42\frac{m}{s^2}

Explanation:

Given the velocity function:

v=0.86\frac{m}{s^3}t^2

you can obtain the instantaneous acceleration "a" as its first derivative:

a=\dot{v}=2\cdot0.86\frac{m}{s^3}\cdot t=1.72\frac{m}{s^3}t

To determine the value of "a" when the velocity was 12m/s, you need to figure out the value for "t" when this happens. At what time t is the velocity 12m/s?

12.0\frac{m}{s}=0.86\frac{m}{s^3}t^2\implies t=3.74s

This value of t is less than the 5 seconds mentioned in the text - so that is a good sign that the formula is valid for this value. And so you can use t=3.47s in the derivative (acceleration) above:

a=1.72t=1.72\frac{m}{s^3}\cdot 3.74s = 6.42\frac{m}{s^2}

3 0
3 years ago
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