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2 years ago

Can mechanical energy be used to do work?

Physics

1 answer:

Lady_Fox [76]2 years ago

7 0

Answer:

Mechanical Energy as the Ability to Do Work An object that possesses mechanical energy is able to do work. In fact mechanical energy is often defined as the ability to do work. Any object that possesses mechanical energy.

Explanation:

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Calculate the Energy (E) in joules for that wavelength and record it in the table below. Remember that E = HF, where h the Planc

Viefleur [7K]

In that formula for Energy, 'F' is the frequency of the photon.
But <u>Frequency = (speed)/(wavelength)</u>, so we can write the
Energy formula as
   E = h c / (wavelength) .

So the energy, in joules, of a photon with that wavelength, is . . .

   E = (6.6 x 10⁻³⁴) x (3 x10⁸) / (that wavelength)

      = <em>(1.989 x 10⁻²⁵) / (that wavelength, in meters) .</em>

5 0

2 years ago

A nuclear accident (intentional or unintentional) can cause significant harm to those living nearby or at a distance. Harmful le

MrRissso [65]

Answer:Explained

Explanation:

Invisible gamma radiation does not transmit through Air or through contact thus standard precautions can encompasses more than person-to-person contamination.

Some Standard Precautions are

Hand Hygiene

Use Alcohol based hand disinfectant and always use gloves.

Use Eye shields,Gowns,face shields .

3 0

3 years ago

2 Points

Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

$:\implies$ Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

$:\implies$ 50 N = 10 kg × Acceleration

$:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}$

$:\implies$ Acceleration of the Object = 5 m/s²

4 0

3 years ago

A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r

katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0

3 years ago

Explain why you can only change one variable in an experiment

ohaa [14]

The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to. But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see.

Or maybe BOTH of them working together caused it. You don't know.

So your experiment is not really much good. You need to do it again.

5 0

3 years ago