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3 years ago

How is it possible for two neutral objects to become charged by rubbing them together?

2 answers:

Volgvan3 years ago

8 0

It is possible for neutral objects to become charged by rubbing them together due to friction. Electrons of one object transfer to the other resulting to more electrons in the object than its protons therefore leaving it as charged. Hope this answers the question.

KatRina [158]3 years ago

3 0

the answer is B on plato

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Two charged point particle are located at two vertices of an equilateral triangle and the electric field is zero at the third ve

Debora [2.8K]

Answer:

Option E

Explanation:

In the presence of two point charges at the two vertices of an equilateral triangle, the resultant electric field at the third vertex due to these charges can not be zero whether the charges are identical or not.

The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.

This can only be achieved if another charge is present.

4 0

4 years ago

At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with

MAVERICK [17]

Answer:

first one is b 2nd one is a 3rd is c and the 4th one is c also

Explanation: have a nice day

6 0

2 years ago

Read 2 more answers

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

3 years ago

In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?

KengaRu [80]

Answer:

3 x 10^5 J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5 J

5 0

4 years ago

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

dolphi86 [110]

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

5 0

3 years ago