What was the name of the first steam-powered locomotive that was developed by george stephenson?

A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

On substituting the given values, we get

⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

8 0

3 years ago

True or false if the temperature does not change but the mass of the object increases the thermal energy in the object decreases

Kay [80]

Answer: im pretty sure its true

6 0

2 years ago

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:

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

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

So

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

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

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

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

7 0

3 years ago

What is the average speed of an object that travels 6 meters in 2 seconds and then travels 3 meters in 1 second?

olya-2409 [2.1K]

Answer:

3 m/s

Explanation:

Average Speed = $\frac{Total Distance}{Total time}$

Plug in the numbers, it will be (6m + 3m) divided by (2s + 1s), which is 9m/3s, which equals to 3m/s.

4 0

2 years ago

When does acceleration due to gravity equal 9.8 m/s downward?

Elza [17]

Answer:

b

Explanation:

i took the quiz i think its right

7 0

2 years ago