A stock person at the local grocery store has a job consisting of the following five segments:1) picking up boxes of tomatoes from the stockroom floor

2)accelerating to a comfortable speed.

3) Carring the boxes to the tomato display at constant speed.

4)decelerating to a stop.

5) lowering the boxes slowly to the floor.

During which of the five segments of the job does the stock person do positive work on the boxes?

A) (2) and (3)

B(1) and (2)

C) (1) only

D) (1), (2), (4) and (5)

E) (1) and (5)

Answers

Answer 1

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answers

Answer:

a). $E_(kA)=1.2635 J$

b). $V_(B)=4.535(m)/(s)$

c). Δ $E_(t)=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J$

b).

$E_(kB)=(1)/(2)*m*v_(B) ^(2)$

$V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)$

c).

net work= EkA+EkB

$E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J$

A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

P = Fv................ Equation 1

Where:

P = work done on the horse in one second
F = Force of the horse
v = Velocity of the wagon

From the question,

Given:

F = 1800 N
v = 0.4 m/s

Substitute these values into equation 1

P = 1800×0.4
P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

Answers

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

Final answer:

The Cartesian components of the resultant force and the couple moment are calculated by summing up all the forces and moments acting on the object. The resultant force is 1724 N and the couple moment is 29.764 N*m.

Explanation:

The resultant force and couple moment in the Cartesian coordinate system can be obtained by summing up all the forces and moments acting on the object. In this case, we have the forces F1, F2, F3, F4 and the couple moment MA acting on the object. The resultant force (FR) can be calculated as the sum of all the forces, i.e., FR = F1 + F2 + F3 + F4. Using the values given, FR = 510 N + 306 N + 501 N + 407 N = 1724 N. The resultant moment (MR) can be calculated as the sum of all the moments, i.e., MR = d1*F1 + d2*F2 + d3*F3 + d4*F4 - MA. Using the values given, MR = 0.880 m * 510 N + 1.11 m * 306 N + 0.560 m * 501 N + 2.08 m * 407 N - 1504 N*m = 29.764 N*m. Therefore, the Cartesian components of the resultant force and the couple moment are 1724 N and 29.764 N*m respectively.

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Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Answers

The spacing between the two slits is 0.221mm.

The spacing between the two slits is given as,

$D=(\lambda L)/(y)$

Where $\lambda$ is wavelength, y is fringe spacing and L is length of screen.

Given that, $\lambda=589nm,L=150cm,y=4mm$

Substitute in above equation.

$D=(589*10^(-9)*150*10^(-2) )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm$

Hence, the spacing between the two slits is 0.221mm.

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A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answers

The acceleration of the box is 0.81 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

$\sum F= ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:

- the horizontal component of the force exerted by the rope, which is equal to

$F_x = F cos\theta = 250*cos 32 = 212 N$

the frictional force, acting in the opposite direction, which is equal to

$F_f = \mu *mg = 171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the

box,

$F_x - F_f = ma\na = 0.81 m/sec^2$

The acceleration of the box is 0.81 m/sec².

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An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?