All categories

2 years ago

Please help someone. A generator consumes 147 joules of energy in fuel and produces 75 joules of energy

Physics

1 answer:

timama [110]2 years ago

7 0

Answer:

Explanation:

Current effiency

= 75/147

= 0.5102

New Efficiency

= 120/100 × 0.5102

= 0.6122

Only option A satisfies this efficiency.

You might be interested in

Which element is the main component of<br> steel and the most widely used of all metals

Blizzard [7]

Answer:

iron

Explanation:

8 0

2 years ago

Read 2 more answers

What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse

Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last

3 0

3 years ago

I dont know what pfp to put on today so I'm asking u guys....

Mandarinka [93]

Answer:

i think number 2 should be your pfp

5 0

3 years ago

Cleo stated that light travels through air in straight paths, and when it moves from air to water, light changes direction, spee

Andru [333]

Answer:

Light slows down when it moves from air into water.

Explanation:

It is the property of light that travels faster in a less dense medium.

In a more dense medium, the speed of the light slows down and bends towards the normal.

The air is less dense medium and water is a more dense medium.

When light passes from air to water, the light bends.

This is known as the refraction of light.

4 0

3 years ago

Read 2 more answers

The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

8 0

3 years ago