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2 years ago

fill in the blank ,if a charge body touches the disc of an uncharged electroscope the leaves ------------------

Physics

1 answer:

Ilia_Sergeevich [38]2 years ago

7 0

Answer:

leaves will deverge

Explanation:

because of the nagwtive charge will become positive

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A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m

Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

$I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2$

The moment of inertia of the rod about its center is calculated as follows;

$I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2$

The initial rotational kinetic energy of the disk and rod;

$K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J$

The final rotational kinetic energy of the disk-rod system is calculated as follows;

$K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J$

The loss in rotational kinetic energy due to the collision is calculated as follows;

$\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J$

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

8 0

2 years ago

The photo shows a system that consists of a person on a diving board. What change could you make to the system that would decrea

Sliva [168]

Answer:

Explanation:

If you decrease the height of the diving board, you will create less energy because you will build up less acceleration jumping from a lower height. Hope this helps ya!

5 0

3 years ago

Read 2 more answers

500km is equal to how many millimeters

Sindrei [870]

Answer:

500000000

if you can give me brainliest that would be great

7 0

4 years ago

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Studentka2010 [4]

Hello!

$\large\boxed{F = 100N}$

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

6 0

3 years ago

On Earth, friction can be _____ but not removed.

Whitepunk [10]

Answer:

reduced

Explanation:

The use of bearing surfaces that are themselves sacrificial, such as low shear materials, of which lead/copper journal bearings are an example

4 0

3 years ago

fill in the blank ,if a charge body touches the disc of an uncharged electroscope the leaves ------------------​

fill in the blank ,if a charge body touches the disc of an uncharged electroscope the leaves ------------------