What is the acceleration of a 10 kg block that experiences a 50 N applied force as it slides across a

In a real machine, the work output is always less than the work input. a. True b. False

Artist 52 [7]

In a real machine, the work output is always less than the work input is true. The answer is letter A. it follows the law of entropy where no energy can be converted completely into work. Under this law, Carnot’s theorem states that Carnot’s engine can perform 100% of work. However, no such engine has ever succeeded the conversion of work into 100%. The greatest efficiency so far is at 80%. Because there will always be factors that could affect the conversion of work.

4 0

3 years ago

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

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A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

8 0

3 years ago

I feel really isolated these days and wish I had more friends. I've chosen to put myself in new social situations to try to make

leva [86]

Behaviorist approach to psychology

3 0

3 years ago

How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?

Vaselesa [24]

Answer:

Time=120seconds

Explanation:

S=300m

V=2.5m/s

t=?

V=S/t

t=S/V

t=300/2.5

t=120 second

7 0

3 years ago

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