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NikAS [45]
2 years ago
5

5. If x = 0.6 and y = 5, the value of 2(xy)^3 is...

Physics
2 answers:
taurus [48]2 years ago
3 0

Explanation:

=2(0.6 * 5)³

=2( 3 )³

=2(27)

=54

VashaNatasha [74]2 years ago
3 0

Answer:

54

Explanation:

2(xy)^3

=2(0,6×5)^3

=54

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3 years ago
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A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal e
Nesterboy [21]

Answer:

 U₂ = 400 KJ      

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ

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3 years ago
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3 years ago
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What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a
Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2

v = r ω

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}

I = \dfrac{m(2gh - v^2)r^2}{v^2}

I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}

I =mr^2(0.287)

I = 0.287 MR²

3 0
3 years ago
7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec
lys-0071 [83]

The time elapsed is 9 seconds

Explanation:

The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:

v=u+at

where :

v is the final velocity  of the ball

u is the initial velocity

a is the acceleration

t is the time  elapsed

For the ball in this problem, we have:

u = 3 m/s is the initial velocity

v = 34.5 m/s is the final velocity

a=3.5m/s^2 is the acceleration

Solving for t, we find the time taken for this change in velocity:

t=\frac{v-u}{a}=\frac{34.5-3}{3.5}=9 s

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