What is the function of a transformer?Why is this done?

Physics

1 answer:

ollegr [7]4 years ago

8 0

Answer:

A transformer is an electrical device for changing the size of an AC voltage

Explanation:

A transformer is an electrical device for changing the size of an alternating current voltage.it acts to increase or decrease the electromotive force of an alternating current

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A 10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the spring, initially displaced 25 cm b

NARA [144]

Answer:

Explanation:

1. Find spring constant k. From a free body diagram, you will get the forces on the 10 kg mass, with a displacement d. It will be gravity pulling the mass down and the spring force pulling the mass up. The 10 kg mass is in equilibrium. The resulting equation will be:

$F = m_1a = kd - m_1g = 0 => m_1g = kd => k = \frac{m_1g}{d}$

2. Use the result from 1. to find the equations of motion. In general they are given by:

$x(t) = Acos(\omega t + \phi), v(t) = -\omega Asin(\omega t + \phi)$ , where ω is: $\omega = \sqrt{\frac{k}{m_2}}=\sqrt{\frac{m_1g}{m_2d}}$

To find the amplitude A and the phase angle Ф, use the given initial conditions:

m₂ = 2 kg, x(0) = -0.25 m, v(0) = 2m/s

$-0.25 = Acos(\phi)\\2 = -\omega Asin(\phi)= -\sqrt{\frac{m_1g}{m_2d}}Asin(\phi)\\-0.24 = Asin(\phi)$

Solving for Ф:

$\frac{0.24}{0.25}=tan(\phi) => \phi = 0.76$

Solving for A:

$-0.25 = Acos(\phi) => A = -\frac{0.25}{cos(\phi)}=-0.35$

The equation for x(t) is now:

$x(t) = -0.35 cos(\sqrt{\frac{m_1g}{m_2d}t} +0.76)$

The frequency f is given by: $f = \frac{\omega}{2\pi}$

The period T is given by: $T = \frac{2\pi}{\omega}$

8 0

3 years ago

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is

Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$