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AfilCa [17]
3 years ago
5

You are carrying a 7.0-kg bag of groceries as you walk at constant velocity along the sidewalk. You walk a distance of 82 meters

and hold the bag at a height of 1.5 m above the sidewalk. How much work against gravity do you do on the bag? Explain.
0 J

100 J
Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

the work againts gravity is zero (0)

Explanation:

This is because the work is always done, by a force that acts parallel in the direction of the displacement of the body.

In this case, the gravity and the weight of the bag is not acting in the direction of the displacement, therefore there is no work done.

You might be interested in
A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of
Doss [256]

Solution :

a). From Newtons second law,

F = ma

The total tension force is 2T.

∴ 2T - (m + M)g = (m+ M)a

Then

$a=\frac{2T-(m+M)g}{m+M}$

$a=\frac{2\times 600-(52+63)9.8}{52+63}$

   $=0.63 \ m/s^2$

b). From the person,

   F = ma

 T - Mg + N = Ma

or N = Ma + Mg - T

        = (63 x 9.8) + (52 x 9.8) - 600

        = 617.4 + 509.6 - 600

        = 527 N

6 0
3 years ago
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at
GuDViN [60]

Answer:

a) v = 11.24 m / s ,    θ = 17.76º   b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

     p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

      p_{x f}= (m₁ + m₂) vx

       p₀ₓ =  p_{x f}

       m₁ v₁ = (m₁ + m₂) vₓ

     vₓ = m1 / (m₁ + m₂) v₁

let's calculate

       vₓ = - 900 / (900 + 1200) 25

       vₓ = - 10.7 m / s

Axis y

initial moment

      p_{oy}= m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

     p_{fy}= (m₁ + m₂) v_{y}

     p_{oy} = p_{fy}

     m₂ v₂ = (m₁ + m₂) v_{y}

     v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

    v_{y} = 1200 / (900+ 1200) 6

    v_{y}  = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

      v = √ (vₓ² + v_{y}²)

      v = RA (10.7²2 + 3,428²2)

      v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

      tan θ = v_{y} / Vₓ

      θ = tan-1 v_{y} / vₓ)

      θ = tan -1 (3,428 / 10.7)

       θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

      Kf = 1/2 (m1 + m2) v2

      K₀ = ½ m₁ v₁² + ½ m₂ v₂²

      Kf = ½ (900 + 1200) 11.24 2

      Kf = 1.3265 105 J

      K₀ = ½ 900 25²  + ½ 1200 6²

      K₀ = 2,8125 10⁵ + 2,16 10₅4

        K₀ = 3.0285 105J

the wasted energy is

        Kf / K₀ = 1.3265 105 / 3.0285 105

        Kf / K₀ = 0.4380

         

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

5 0
3 years ago
A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a
Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}

From Newton's second law

F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}

Squarring both sides

112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m

The height from which the student fell is 5.7141 m

5 0
3 years ago
The Chartered Institute of Management Accountants defines activity-based cost management as follows: "An approach to the costing
egoroff_w [7]

Answer:

ABC is the activity based accounting. It is the costing done for each separate activity which maybe unit level, batch level, product sustaining or facility sustaining.

Explanation:

The basic difference between ABC and traditional costing systems can be explained with the help of the following diagram.

Traditional Costing System

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation

  1. Cost Centers ( normally Departments)
  2. Cost Centers ( normally Departments)

N.  Cost Centers ( normally Departments)

Second Stage Allocation ( Direct Labor Or Machine Hours)

Cost Objects ( products, services and customers)

ACTIVITY BASED COSTING SYSTEMS

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation ( resource cost drivers)

  1. Activity Cost Centers
  2. Activity Cost Centers

N. Activity  Cost Centers

Second Stage Allocation ( activity cost drivers)

Cost Objects ( products, services and customers) (Direct Costs)

Four steps are involved in the design of ABC systems.

  1. identifying the major activities that take place in an organization
  2. assigning cost to cost pools / cost centers for each activity
  3. determining the cost for each major activity
  4. assigning the cost of activities to products according to the product's demand for activities.

The first two steps relate to the first stage and final two steps to the second stage of the two - stage allocation process shown above.

5 0
3 years ago
Anaerobic metabolism:
goblinko [34]

Explanation:

B. leads to muscle strain.

5 0
3 years ago
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