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Mamont248 [21]
3 years ago
6

Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horiz

ontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.
Physics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

a) a = 2.383 m / s², b)   T₂ = 120,617 N , c)   T₃ = 72,957 N

Explanation:

This is an exercise of Newton's second law let's fix a horizontal frame of reference

in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.

a) request the acceleration of the system

we can take the sledges together and write Newton's second law

     T = (m₁ + m₂ + m₃) a

    a = T / (m₁ + m₂ + m₃)

     a = 143 / (10 +20 +30)

     a = 2.383 m / s²

b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg

on the sled of m₁ = 10 kg

          T - T₂ = m₁ a

in this case T₂ is the cable tension

           T₂ = T - m₁ a

            T₂ = 143 - 10 2,383

            T₂ = 120,617 N

c) The cable tension between the masses of 20 and 30 kg

            T₂ - T₃ = m₂ a

             T₃ = T₂ -m₂ a

             T₃ = 120,617 - 20 2,383

             T₃ = 72,957 N

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4 years ago
Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87
marshall27 [118]

Answer:

a)  \lambda=0.935\ \textup{m}

b) f=36.19\approx 36\ \textup{Hz}

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

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

where,

\lambda = wavelength

on substituting the values, we get

1.87= 4\times \frac{\lambda}{2}

or

\lambda=0.935\ \textup{m}

b) Speed of the wave (v) in the string is given as:

v =f\lambda

also,

v=\sqrt\frac{T}{(\frac{m}{L})}

equating both the formula for 'v' we get,

f\lambda=\sqrt\frac{T}{(\frac{m}{L})}

on substituting the values, we get

f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}

or

f=\frac{33.84}{0.935}

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3 years ago
Read 2 more answers
The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find
snow_tiger [21]

Answer:

  • The emf of the generator is 6V
  • The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7   + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7   + 0.3r  

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0
3 years ago
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