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
otez555 [7]
3 years ago
11

How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met

ers horizontallyy across the room and place it on a shelf that is 5.7 meters off the ground? Do not type units. Round your answer to te tnths place
Physics
1 answer:
PSYCHO15rus [73]3 years ago
5 0

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

         =  mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

   = 23 x 9.18 x 1

   = 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

   = 23 x 9.18 x 6

   = 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

   = 23 x 9.18 x 5.7

   = 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

                                          = 2681.47

                                          = 2682 (rounding off)

You might be interested in
Layers in Earth's atmosphere are defined as regions in which _____.
Solnce55 [7]

Answer:

b. different types of weather occur

Explanation:

the temperture varies as one moves higher up, the higher one goes the cooler the temperature becomes. So that creates a change in weather condition.

6 0
3 years ago
A car is traveling at a constant velocity of magnitude v0 when the driver notices a garbage can on the road in front of him. at
slava [35]
Total distance travelled by the car is 'd' 
<span>distance trveled before the brakes were applied = v_o * t </span>
<span>distance travld with brakes = d - v_o*t </span>
<span>applying the formula: v^2 - u^2 = -2 a * s </span>
<span>=> 0 - v_o^2 = -2 * a_x * (d- v_o*t) </span>
<span>=> a_x = (v_o^2)/ ( 2 (d-v_o*t)</span>
7 0
3 years ago
Read 2 more answers
By what factor would you need to adjust the length of a pendulum to make the period of 1/2 of what it used to be?
Irina-Kira [14]

Answer:

1/4

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

At half the period:

T/2

= π √(L/g)

= 2π √(L/(4g))

= 2π √((L/4)/g)

So the length would have to be shortened by a factor of 1/4.

3 0
3 years ago
A copper sphere 10 mm in diameter is dropped into a 1-m-deep drum of asphalt. The asphalt has a density of 1150 kg/m3 and a visc
kvasek [131]

Answer:

t = 1964636.542 sec

Explanation:

Given data:

sphere diameter is 10 mm

Density is 1150 kg/m^3

viscosity 105 N s/m^2

We knwo that time taken by sphere can be calculated by following procedure

\tau = \mu \frac{du}{dy}

\frac{F}{A} =  \mu \frac{du}{r}

\frac{\rho_C -\rho_{asphalt} gv}{2 \pi rL} = 10^5 \frac{du}{r}

Solving for du

du = \frac{ (8933 - 1150) 9.81 \frac{4}{3} \pi (10\times 10^{-3})^3}{2\pi \times 1\times 10^5}

du = u = 5.09\times 10^{-7}

u = \frac{1}{t}

t = \frac{1}{5.09\times 10^{-7}} = 1964636.542 sec

6 0
3 years ago
Which is true concerning the acceleration due to gravity? A. It decreases with increasing altitude. B. It is different for diffe
Kisachek [45]

Answer:

The correct answer is  option 'a': It decreases with increase in altitude

Explanation:

Acceleration due to gravity is the acceleration that a  body is subjected to when it is freely dropped from a height from surface of any planet, ignoring the resistance that the object may face in it's motion such as drag due to any fluid.. The acceleration due to gravity is same for all the objects and is independent of their masses, it only depends on the mass of the planet and the radius of the planet on which the object is dropped. it's values varies with:

1) Depth from surface of planet.

2)Height from surface of planet.

3) Latitude of the object.

Hence it neither is a fundamental quantity nor an universal constant.

The variation of acceleration due to gravity with height can be mathematically written as:

g(h)=g_{surface}(1-\frac{2h}{R_{planet}})

where,

R is the radius of the planet

g_{surface} is value of acceleration due to gravity at surface.

hence we can see that upon increase in altitude the value of 'g' goes on decreasing.

5 0
3 years ago
Other questions:
  • When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
    14·1 answer
  • How many electrons are there in 3.5 x 10" C?
    7·1 answer
  • Jimmy held the end of a metal bar over a fire while holding on to the opposite end. After a few minutes, the end he was holding
    6·1 answer
  • Question 11(Multiple Choice Worth 3 points)
    12·1 answer
  • Help!!!! ASAP!!! A loop of area 0.100 m^2 is oriented at 15.5 degree angle to a 0.500 T magnetic field. It rotates until it is a
    5·1 answer
  • A uniform metal rod, with a mass of 2.6 kg and a length of 1.5 m, is attached to a wall by a hinge at its base. A horizontal wir
    10·1 answer
  • What is the relative size and composition of the universe, a galaxy, and a solar system? Is the universe endless?
    12·2 answers
  • A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement
    8·1 answer
  • Explain the basic reason of conduction of an electric current through a conductor.​
    5·2 answers
  • What do you understand by family health?<br>​
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!