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
saw5 [17]
2 years ago
7

Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

angs from the ceiling. Find
(a) the acceleration of the objects and
(b) the tension in the string.
Physics
1 answer:
Oksanka [162]2 years ago
4 0

a) The acceleration of the objects is 3.56 m/s^2

b) The tension in the string is 280.8 N

Explanation:

a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

m_1 g - T = m_1 a (1)

where

m_1 = 45.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

T-m_2 g = m_2 a (2)

where

T is the tension in the string

m_2 = 21.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

T=m_1g - m_1 a

And substituting into (2),

(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g  = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2

b)

From the previous part of the problem we got an expression for the tension in the string:

T=m_1g - m_1 a

Where we have

m_1 = 45.0 kg

g=9.8 m/s^2

a=3.56 m/s^2 is the acceleration, found in part a)

Susbtituting, we find

T=(45.0)(9.8)-(45.0)(3.56)=280.8 N

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

8 0
2 years ago
Read 2 more answers
A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of
diamong [38]

Answer:

 2n t = m λ₀ ,    R = 0.240 mm

Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,

-This phase change is the introduction of a λ/2 change

-The ray passing through the film has a change in wavelength due to the refractive index of the medium

          λ₀ = λ / n

Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is

        2 t = (m’+ ½ + ½) λ₀ / n

        2t = (m’+1) λ₀ / n

         m = m’+ 1

        2n t = m λ₀

        With   m = 0, 1, 2, ...

Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength

The thickness of a hair is the thickness of the film t

           2R = t

             R = t / 2

             R = 0480/2

              R = 0.240 mm

3 0
2 years ago
The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
kifflom [539]
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s
6 0
2 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

ωf = ω₀+ α*t  Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
Other questions:
  • If steam enters a turbine at 600K and is exhausted at 400K, calculate the efficiency of the engine.
    13·2 answers
  • A tennis ball bounces on the floor three times, and each time it loses 23.0% of its energy due to heating. how high does it boun
    9·1 answer
  • A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?
    13·2 answers
  • The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the
    12·1 answer
  • A small first-aid kit is dropped by a rock climber who is descending steadily at 1.3 m/s. After 2.5 s, what is the velocity of t
    12·1 answer
  • A liquid in a test tube has a curved surface such that the edges touching the glass are higher than the surface at the center. T
    5·1 answer
  • Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
    15·1 answer
  • A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
    10·1 answer
  • What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light i
    5·1 answer
  • The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an obj
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!