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
Aleks [24]
3 years ago
7

Hari drag a load of 60 kg along a distance of 12 metre what amount of work does he do also mention the type of work​

Physics
1 answer:
Vlada [557]3 years ago
6 0

Answer:

W = 7056  J

Explanation:

Given that,

Mass of a load, m = 60 kg

Distance moved, d = 12 m

We need to find the amount of work. The work done by an object is given by :

W = Fd

So,

W = mg×d

W = 60 kg × 9.8 m/s²×12 m

W = 7056  J

So, the amount of work done is 7056  J.

You might be interested in
How can sand dunes be re-built?<br><br> PLEASE HELP ME
quester [9]
The basic steps are simple but careful planning is needed. Sand dune restoration should be designed to create a dune that matches the existing natural dune pattern in the area. You can help speed up nature's work by using sand fences and dune plants to collect sand more rapidly. Or in a different way, wind would bring more sand to create a dune. hope this helped, have an amazing day :)
7 0
4 years ago
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp
Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

t = \frac{{2v\sin \theta }}{g}

Substitute 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}} for g and 60^\circ  for \theta in above equation.

t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\  

Spitted water will travel 0.80{\rm{ m}} horizontally.

Displacement of water in this time period is

x = vt\cos \theta

Substitute \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2} for t\rm 60^\circ[tex] for [tex]\theta and 0.80{\rm{ m}} for x in above equation.

\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

{v_v} = v\sin \theta

Substitute 3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}} for v and 60^\circ  for \theta in above equation.

\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

Horizontal component of the velocity is,

{v_h} = v\cos \theta

Substitute 3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}} for v and 60^\circ  for \theta in above equation.

\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

When horizontal ({0.60\;{\rm{m}}} distance away from the fish.  

The time of flight for distance (d) is ,

t = \frac{d}{{{v_h}}}

Substitute 0.60\;{\rm{m}} for d and 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}} for {v_h} in equation t = \frac{d}{{{v_h}}}

\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\

Distance of the insect above the surface is,

s = {v_v}t + \frac{1}{2}g{t^2}

Substitute 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}} for {v_v} and 0.3987{\rm{ s}} for t and - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}} for g in above equation.

\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\

7 0
3 years ago
How many mL in 53,246 Liters, convert using dimensional analysis
satela [25.4K]
A). Both. ... b). Series. ... c). Parallel. ... d). and e). Series.
7 0
3 years ago
A rock at the edge of a cliff has what kind of energy?
Marina CMI [18]
A rock at the edge of a cliff has potential energy because it isn't in motion, but is located in a place that motion could occur in the near future.
7 0
4 years ago
Isla made ice by putting water in the freezer. Freezing is an example of (2 points)
SSSSS [86.1K]

Answer:

a physical change

6 0
3 years ago
Read 2 more answers
Other questions:
  • What is the gravitational force exerted on an object
    15·1 answer
  • What are the two types of modern wind turbines?<br><br> (please list TWO modern wind turbines)
    10·1 answer
  • Use the diagram to help answer these questions.
    15·1 answer
  • The rare earth elements include
    7·1 answer
  • A change in the size or shape of a substance is a ___________ change. physical, chemical, or nuclear?
    5·1 answer
  • How does a toaster oven use micro waves
    11·1 answer
  • One ball is dropped vertically from a window. At the same instant, a second ball is thrown horizontally from the same window.
    9·1 answer
  • What do we mean by "Ohmic"?
    6·1 answer
  • a 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m
    7·1 answer
  • Sending food to a country that has been hit by a hurricane is an example of which foreign policy tool?
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!