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2 years ago

8

The formula for calculating power is work divided by time (power = work ÷ time). What are two ways of stating the same relations

hip?

2 answers:

UNO [17]2 years ago

7 0

If F is constant one could rewrite the equations like this also:

$P = \frac{W}{t} = \frac{F s}{t} = Fv$

nirvana33 [79]2 years ago

4 0

I think they just want you to rewrite it two other ways:

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What happens when an atoms losses an electron-

inysia [295]

It becomes positive

8 0

1 year ago

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac = q ( Va - Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]

= Wac - Wab

= Wac + 3Wac

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4 0

2 years ago

The acceleration of an object is constant when its velocity is :

emmasim [6.3K]

Answer:

B. changing by a constant amount each second

Explanation:

thats my answer

5 0

2 years ago

Read 2 more answers

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

6 0

2 years ago

Energy that cannot be used to do useful work is referred to as Select one: a. potential energy. b. entropy. c. kinetic energy. d

topjm [15]

In thermodynamics, entropy (symbolized as S) is a physical magnitude for a thermodynamic system in equilibrium. It measures the number of microstates compatible with the equilibrium macrostate, it can also be said that it is the reason for an increase between internal energy versus an increase in system temperature.

The universe tends to distribute energy evenly; that is, to maximize entropy. Intuitively, entropy is a physical quantity that, by calculation, allows us to determine the part of energy per unit of temperature that cannot be used to produce work.

Therefore the correct answer is B.

5 0

2 years ago