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3 years ago

Thermal energy produced by friction cannot usually be used to do work. true/false

Physics

2 answers:

Dmitriy789 [7]3 years ago

7 0

Answer:

false

Explanation:

Work may be done in positive or negative direction. When a work is done in positive direction, it is called useful work, when work is done in negative direction, it is known as useless work. Any energy associated with friction produces useless work. That is, the work is done in negative direction.

Therefore, thermal energy produced by friction can be used to do work but in negative direction.

The correct option is false.

maria [59]3 years ago

4 0

I believe the correct response would be true, thermal energy or heat that is produced by friction usually cannot be used to do work.

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tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

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3 years ago

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be

Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

$v_1=\dfrac{72.42 }{2.5}=28.96\ km/h$

Average speed of Prav,

$v_2=\dfrac{119}{1.25}=95.2\ km/h$

Difference in average speed of Ajoba and Prav is :

$v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h$

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

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2 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

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