All categories

3 years ago

How long would it take a 500. W electric motor to do 15010 J of work?

Physics

1 answer:

KIM [24]3 years ago

8 0

Time = energy / power = 15010 / 500 = .... seconds

You might be interested in

Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

$36=\frac{1}{2}m(12)^2$ and isolating the m:

$m=\frac{2(36)}{12^2}$ which gives us

m = .50 kg

3 0

3 years ago

All of the following are examples of electric current except _______.

Aleks04 [339]

<span>Electrical discharge from a charged object
is your answer</span>

4 0

3 years ago

2. The substances that undergo change in a chemical reaction are called

AVprozaik [17]

Reactants

In a chemical reaction, the substances that undergo change are called reactants. The new substances formed as a result of that change are called products.

4 0

3 years ago

A graph can be described as:

pantera1 [17]

D all of the above

....

6 0

3 years ago

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

andrew-mc [135]

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm$

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0

3 years ago