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
NikAS [45]
2 years ago
6

A rope is tied to a box and used to pull the box 2.3 m along a horizontal floor. The rope makes an angle of 30∘ with the horizon

tal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N
Part A

How much work does gravity do on the box?

Part B

How much work does the tension in the rope do on the box?

Part C

How much work does the friction do on the box?

Part D

How much work does the normal force do on the box?

Part E

What is the total work done on the box?
Physics
1 answer:
Savatey [412]2 years ago
6 0

Answer:

Answered

Explanation:

A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.

W= FS cosθ

θ= 90 ⇒cos90 = 0 ⇒W= 0

B) work done by tension

W= Tcosθ×S= 5cos30×2.30= 10J

C) Work done by friction force

W= f×s=1×2.30= 2.30 J

D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.

E) The net work done= Work done by tension in the rope - frictional work

=10-2.30= 7.7 J

You might be interested in
CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk
malfutka [58]

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

5 0
3 years ago
PLEASE PROVIDE AN EXPLANATION.<br><br> THANKS!!!
ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(c) 1250 N

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

(c) The tension is:

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))

y = 0.0365 m

8 0
2 years ago
Read 2 more answers
Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te
zlopas [31]

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

4 0
3 years ago
A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t
pshichka [43]

Answer:

<h2>The amount of torque put on the car is 33,000Nm</h2>

Explanation:

Formula for calculating torque is expressed as T = rFsin\theta\\ where;

r is the radius of the  of the arm of the jack = 3m

F is the force exerted = 11000

\theta\\ is the angle of rotation = 90°

On substituting;

T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm

6 0
3 years ago
Evan drew a diagram to illustrate radiation.
Vesna [10]

The correct answer is:

D. Electromagnetic waves.

The arrows represent electromagnetic waves.

|Huntrw6|

8 0
3 years ago
Other questions:
  • How many protons are in the radioactive isotope 40/19K?
    14·1 answer
  • Which season is occurring in earths northern hemisphere when earths Southern Hemisphere is tilted toward the sun
    9·2 answers
  • How many work is done when a force of 33n pulls wagon 13meters
    5·1 answer
  • 7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field
    6·1 answer
  • to travel at a constant velocity, a car exerts force that is 551N to balance air resistance. how much work does the car do on th
    9·1 answer
  • 1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane
    13·1 answer
  • What is released through nuclear fusion in stars?<br><br> energy<br> gas<br> mass<br> pressure
    9·2 answers
  • How can you put a lever into a rube goldberg machine? answer ASAP
    8·2 answers
  • A 1000 kg boulder sits at the top of a 50-meter high cliff. Determine the amount of potential energy possessed by the boulder if
    15·2 answers
  • A plane accelerates to a velocity of 240 m/s in 11 s by which time it had traveled 1,400 m down the runway what were it average
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!