answer.
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
liraira [26]
2 years ago
14

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t

ravels 2.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at a 60.0-degree angle, calculate the work done on the bag by the following:
(a) The force of gravity
(b) The normal force
(c) The friction force
(d) The conveyor belt
(e) The net force.
Physics
1 answer:
LenKa [72]2 years ago
4 0

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

You might be interested in
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour
slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0
2 years ago
In general, how do you find the average velocity of any object falling in a vacuum?
irina [24]
In general, how do you find the average velocity of any object falling in a vacuum? (Assume you know the final velocity.) Multiply the final velocity by final time. 3. Calculate : Distance, average velocity, and time are related by the equation, d = v • t A
5 0
1 year ago
Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com
Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

7 0
1 year ago
A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular vel
KengaRu [80]

Answer:

The net torque is 0.0372 N m.

Explanation:

A rotational body with constant angular acceleration satisfies the kinematic equation:

\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta (1)

with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}}

The negative sign indicates the sphere is slowing down as we expected.

Now with the angular acceleration we can use Newton's second law:

\sum\overrightarrow{\tau}=I\overrightarrow{\alpha} (2)

with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:

I = \frac{2MR^{2}}{5}

With M the mass of the sphere an R its radius, then:

I = \frac{2(1.85)(\frac{0.45}{2})^{2}}{5}=0.037 kg*m^2

Then (2) is:

\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm

7 0
2 years ago
Read 2 more answers
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
Other questions:
  • The first thing to focus on when creating a workout plan is
    7·2 answers
  • A scientist measures the growth of a bamboo plant over time. The table below shows the results.
    5·2 answers
  • You throw a beanbag in the air and catch it 2.2 s later at the same place at which you threw it. How high did it go? What was th
    9·1 answer
  • Jack (mass 52.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. he collides with jill (mass 49.0 k
    9·1 answer
  • The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.
    13·1 answer
  • Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st
    7·1 answer
  • A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase w
    8·1 answer
  • An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut
    8·1 answer
  • A propeller blade at rest starts to rotate from t = 0 s to t = 5.0 s with a tangential acceleration of the tip of the blade at 3
    6·1 answer
  • Two objects interact with each other and with no other objects. Initially object A has a speed of 5 m/s and object B has a speed
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!