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4 years ago

Reducing friction in a machine

Physics

1 answer:

SVETLANKA909090 [29]4 years ago

7 0

Isn't the correct answer it increases efficiency, so C?

Sorry if I'm wrong but pretty sure that's right!

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A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

ruslelena [56]

Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

$r_{B} = \frac{1}{0.299} = 3.34\ m$

5 0

4 years ago

Which of the following is part of a atom

finlep [7]

Atoms are made of neutrons, electrons and a proton.

8 0

3 years ago

Read 2 more answers

What is the length of an aluminum rod at 65Â°C if its length at 15Â°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2

Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

$\Delta L=L\alpha \Delta T$

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

$\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m$

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0

3 years ago

Read 2 more answers

Will give brainiest to best answer

Gala2k [10]

Answer:

Explanation:

6 0

3 years ago

Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why

Morgarella [4.7K]

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f = 2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

5 0

3 years ago