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3 years ago

Deduce the diamension of work

Physics

2 answers:

djyliett [7]3 years ago

7 0

Answer:

Explanation:Quantity

Dimension

Unit

SI Symbol

Formula

Base units:

length

[L]

meter

mass

[M]

kilogram

temperature

[Q]

kelvin

time

[T]

second

Derived units:

area

[L2]

square meter

volume

[L3]

cubic meter

velocity

[L T-1]

meter per second

m s-1

acceleration

[L T-2]

meter per second squared

m s-2

density

[M L-3]

kilogram per cubic meter

kg m-3

force

[M L T-2]

newton

kg m s-2

pressure

[M L-1 T-2]

pascal

N m-2

stress

[M L-1 T-2]

pascal

N m-2

energy

[M L2 T-2]

joule

N·m

quantity of heat

[M L2 T-2]

joule

N·m

work

[M L2 T-2]

joule

N·m

power

[M L2 T-3]

watt

J s-1

viscosity, dynamic

[M L-1 T-1]

pascal-second

Pa·s

viscosity, kinematic

[L2 T-1]

square meter per second

m2 s-1

specific heat

[L2 Q-1 T-2]

joule per kilogram-kelvin

J kg-1 K-

Mademuasel [1]3 years ago

4 0

Answer:

Unit of work done is Joule. Therefore, the dimension formula of work is represented as [M1 L2 T-2].Explanation:

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

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Deduce the diamension of work​

Deduce the diamension of work