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
Salsk061 [2.6K]
3 years ago
14

The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu

lley are massless and that there is no friction in the pulley. If the masses have the values m1=21.1 kg and m2=12.9 kg, find the magnitude of their acceleration ???? and the tension T in the rope. Use ????=9.81 m/s2.

Physics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

a=2.36\ m/s^2

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass  less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

m_1g-m_2g=(m_1+m_2)a

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

a=2.36\ m/s^2

Lets tension in string is T

m_1g-T=m_1a

T=m_1(g-a)

T=21.1(9.81-2.36) N

T=157.06 N

You might be interested in
What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across
enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0
3 years ago
Read 2 more answers
A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
3 years ago
While driving down his street one evening, Jonah notices that his neighbor has laid out an electric fan for the garbage to pick
Kryger [21]

Answer: I have no idea either i need help aswell

Explanation:

4 0
3 years ago
Read 2 more answers
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
Which undergoes greater acceleration: an airplane that goes from 1000 km/h to 1005 km/h in 10 seconds or a skateboard that
Alex_Xolod [135]

Answer:

Skateboard

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

The airplane's acceleration is:

a = (1005 km/h − 1000 km/h) / 10 s

a = 0.5 km/h/s

The skateboard's acceleration is:

a = (5 km/h − 0 km/h) / 1 s

a = 5 km/h/s

6 0
3 years ago
Other questions:
  • A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
    13·2 answers
  • A satellite orbits the earth a distance of 2.50 × 10^7 m above the planet’s surface and takes 6.43 hours for each revolution abo
    8·1 answer
  • In which state of matter are particles spread farthest apart from another?
    7·2 answers
  • Scientists described light as a wave because the results of many experiments with light demonstrated that light behaves as a wav
    6·2 answers
  • The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^
    10·2 answers
  • A baseball leaves a bat and flies upward and toward center field. Do not ignore air resistance. What forces are exerted on the b
    11·1 answer
  • A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
    14·1 answer
  • For each pair of vehicles, choose the vehicle that would experience the greatest force of impact in a collision. Type your answe
    10·1 answer
  • An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
    15·1 answer
  • Hi guys <br> I suggest to make group to exchange the informations if they need add their name here
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!