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2 years ago

Fizik

Physics

1 answer:

vekshin12 years ago

3 0

Answer:

P = 24.32 W

Explanation:

The question is, "Two resistors 5 Ω and 10 Ω are connected in parallel with a 9 V power supply. Calculate the output power of the power supply.".

The voltage of the power supply, V = 9 V

Resistor 1, R₁ = 5 Ω

R₂ = 10 Ω

The equivalent of parallel combination of resistors is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_{eq}}=\dfrac{1}{5}+\dfrac{1}{10}\\\\R_{eq}=3.33\ \Omega$

The power of the output is given by :

$P=\dfrac{V^2}{R_{eq}}\\\\=\dfrac{9^2}{3.33}\\\\P=24.32\ W$

So, the output power is equal to 24.32 W.

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A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

$1 T : 4 A = t : 2 A$

and solving for t we find

$t=\frac{(1T)(2 A)}{4A}=0.5 T$

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

$1 T : 4 A = t : 5 A$

And by solving for t, we find

$t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T$

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2 years ago

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

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Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

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3 years ago

A current of 3.75 A in a long, straight wire produces a magnetic field of 2.61 Î¼T at a certain distance from the wire. Find thi

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Given :

Current, I = 3.75 A .

Magnetic Field, $B = 2.61\times 10^{-4}\ T$

To Find :

The distance from the wire.

Solution :

We know,

$B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m$

Hence, this is the required solution.

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2 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

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2 years ago

A bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go

Umnica [9.8K]

Answer:

20 m

Explanation:

4.00 m/s x 5 seconds

4 m/s x 5 s = 20m

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3 years ago

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