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3 years ago

Is there any possibility to make 100% efficient system

Physics

1 answer:

Ganezh [65]3 years ago

4 0

Answer:

yes

Explanation:

its awesome

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Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a const

Dafna1 [17]

Answer:

$\frac{dA}{dt} = 188.5 m^2/s$

Explanation:

As we know that area of the circle at any instant of time is given as

$A = \pi r^2$

now in order to find the rate of change in area we will have

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

here we know that

rate of change of radius is given as

$\frac{dr}{dt}= 1 m/s$

radius of the circle is given as

$r = 30 m$

now we have

$\frac{dA}{dt} = 2\pi (30)(1)$

$\frac{dA}{dt} = 60\pi$

$\frac{dA}{dt} = 188.5 m^2/s$

8 0

4 years ago

How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?

quester [9]

Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

$W=q\Delta V$

where

W is the work done

q is the charge

$\Delta V$ is the potential difference

In this problem, we have

q = 0.0050 C is the charge

$\Delta V=9.0 V$ is the potential difference

Using the formula, we find the work done:

$W=(0.0050 C)(9.0 V)=0.045 J$

4 0

3 years ago

Read 2 more answers

Images formed by a convex mirror are always

gladu [14]

Answer:

Images formed by a convex mirror are always virtual

Explanation:

A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.

<u>OAmalOHopeO</u>

4 0

3 years ago

Determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilo

FrozenT [24]

Answer:

The average speed of the Earth in its orbit is 4.74 km/s.

Explanation:

Distance between the Earth and the Sun ,r= 149.60 million km = $149.60\times 10^6 km$