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Arturiano [62]
3 years ago
9

How much work is done lifting a 12-m chain that is initially coiled on the ground and has a density 2 kg/m so that its top end i

s 11 m above the ground? (Assume that acceleration due to gravity is g = 9.8 m/s2.)
Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

2587.2 J.

Explanation:

From potential energy,

The work done to lift the chain = potential energy of the chain.

W = mgh............... Equation 1

Where W = work done to lift the chain, m = mass of the chain, g = acceleration due to gravity of the chain, h = height of the chain.

But,

m = m'd............... Equation 2

Where m' = density of the chain, d = length of the chain.

Substitute equation 2 into equation 1

W = m'dgh................ Equation 3

Given: m' = 2 kg/m, d = 12 m, h = 11 m, g = 9.8 m/s²

Substitute into equation 3

W = 2(12)(11)(9.8)

W = 2587.2 J.

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Opposites attract and when you stick a negatively charged objects to positively charged objects, they tend to stick together. When you pick up electrons, it increases the number of electrons which will make the object negatively charged. 

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5 0
2 years ago
The Zero Gravity Research Facility at the NASA Glenn Research Center includes a
denis23 [38]

Answer: (a) t = 5.44 sec

(b) vf = 53.31 m/s

(c) s = 5.0m

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

  t² = 145/4.9

   t² = 29.59

    t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

   a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

4 0
3 years ago
7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2
alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

\displaystyle a = \frac{\Sigma F}{m},

where

  • a is the acceleration of the object in \text{m}\cdot\text{s}^{-2},
  • \Sigma F is the net force on the object in Newtons, and
  • m is the mass of the object in kilograms.

As a result,

\displaystyle m = \frac{\Sigma F}{a}.

Assume that all other forces on this object are balanced. The net force on the object will be 100\;\text{N}. The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}.

\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}.

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}.

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