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dimaraw [331]
3 years ago
7

A 60​-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

chain has a density of 10 ​kg/m. Use 9.8 m divided by s squared for the acceleration due to gravity. a. How much work is required to wind the entire chain onto the cylinder using the​ winch? b. How much work is required to wind the chain onto the cylinder if a 35​-kg block is attached to the end of the​ chain?
Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

  • Density of the chain = \rho\ =\ 10\ kg/m.
  • Length of the chain = L = 60 m
  • Acceleration due to gravity = g = 9.81 m/s^2

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = \rho dy

Work done due to the small element,

dw\ =\ \rho g (60\ -\ y)dy\\

Total work done to wind the entire chain = w

w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow  w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J

part (b)

  • mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J

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A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
An object at rest starts accelerating.
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Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0
3 years ago
A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

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What is a small body that follows a highly elliptical orbit around the sun
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A Planet, such as (pluto)
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A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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#LearnwithBrainly

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