1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Crank
4 years ago
13

How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if t

he effective coefficient of friction was 0.50
Physics
1 answer:
-BARSIC- [3]4 years ago
7 0

Answer:

7.1 J

Explanation:

From the question,

Work done by the mover  = work done in pushing the crate + work done against friction

W = W'+Wf................. Equation 1

W = mgd+mgμd............ Equation 2

W = mgd(1+μ)................ Equation 3

Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of  friction.

Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5

constant: g = 9.8 m/s²

Substitute these values into equation 3

W = 46×9.8×0.0105(1+0.5)

W = 7.1 J

You might be interested in
How do the prefixes micro,<br> nano and pico relate to each<br> other?
In-s [12.5K]

Answer:

because they are same and their properties

8 0
3 years ago
How might viral diseases be prevented?
Sindrei [870]

Answer:

There are antiviral medicines to treat some viral infections. Vaccines can help prevent you from getting many viral diseases.

3 0
3 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2&#10;}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
3 years ago
A heavenly body in another's shadow is ____
Alex_Xolod [135]
<span>It is called a lunar eclipse</span>
8 0
4 years ago
Read 2 more answers
Which form of
Soloha48 [4]
I think it’s C b/c it works for me
3 0
3 years ago
Other questions:
  • Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellit
    8·1 answer
  • If you use the same force on a less massive object what happens to the acceleration
    13·1 answer
  • What's the relationship between geology and the biosphere
    9·1 answer
  • A 600-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 200-g mass traveling toward it also at 8.0 m/s. W
    9·1 answer
  • Can someone help me with this real quick?
    13·1 answer
  • A jet airplane is in level flight. The mass of the airplane is m=8930 kg. The airplane travels at a constant speed around a circ
    5·1 answer
  • Determine the momentum of a 15,000 kg truck traveling at 15 m/s
    14·1 answer
  • Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r
    8·1 answer
  • Ava runs track and field and is a long-distance runner. She has a big race coming up and needs to eat food that will provide her
    9·2 answers
  • A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0 s, the x components of the puck's i
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!