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3 years ago

What is the use of 'ground' in electric circuit?

1 answer:

5 0

In electrical engineering, ground or earth is the reference point in an electrical circuit from which voltages are measured, a common return path for electric current, or a direct physical connection to the earth. Electrical circuits may be connected to ground (earth) for several reasons.

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago

A ball thrown with 50N of force accelerates at 25 m/s2. What is the mass of the ball?

MrRissso [65]

F=ma
Force is 50N. Acceleration is 25 m/s^2.
50N=m*25 m/s^2
Divide both sides by 25.
mass=2 kg

7 0

3 years ago

How are mountain ecosystems different from other ecosystems?

Andreyy89

Answer:

They host three or more distinct ecosystems at different elevations.

Explanation:

I checked it, and its right.

3 0

2 years ago

The vertical component of the projectile motion of an object depends on which of these? initial velocity or angel of trajectory

SVEN [57.7K]

It depends on both of them.

In fact, the projectile begins its motion with an initial velocity of $v_0$ and an angle of $\alpha$ . On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by
$v_y (t) = v_0 sin \alpha + gt$
and as it can be seen, this depends on both initial velocity and angle.

3 0

3 years ago