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Andrew [12]
3 years ago
9

What is the difference between work and power. will mark brainliest

Physics
2 answers:
katen-ka-za [31]3 years ago
7 0

Answer:

Work is defined as the process of energy transfer to the motion of an object through the application of force. This is usually represented as the product of force and displacement. The SI unit of work is Joule.

Explanation:

Power is defined as the amount of energy transferred in unit time. The SI unit of power is the watt. One watt is equal to one joule per second. Power is a scalar quantity.

Ipatiy [6.2K]3 years ago
4 0

I believe

Work is the method of the conversion of energy by the application of force to the motion of an object. Power is the unit-time sum of energy transmitted.

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A is the only answer because

the day of light and the rat of death starved with pyramids, so this We will all die
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Cu ce viteza s-a deplasat un vehicul daca la ora 16:45 se afla in dreptul bornei kilometrice 169 iar la ora 17:50 se afla la bor
jolli1 [7]

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A 9 volt battery produces a current of 0.2A. What is the resistance?
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4 years ago
If the planes of a crystal are 3.50 (1 A= 10^-10m = Ångstrom unit) apart, what wavelength of electromagnetic waves are needed so
goldenfox [79]

Answer:

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

Explanation:

We can use Bragg's Law's equation to solve this problem. The Bragg's Law's equation is written as follows:

mλ = 2d Sin θ

where,

m = order of reflection = 1

λ = wavelength = ?

d = distance between the planes of crystal = 3.5 x 10⁻¹⁰ m

θ = strike angle of waves on plane = 22°

Therefore, substituting the respective values in the equation, we get:

(1)λ = (2)(3.5 x 10⁻¹⁰ m)(Sin 22°)

<u>λ = 2.62 x 10⁻¹⁰ m = 0.262 nm</u>

4 0
3 years ago
A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0
3 years ago
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