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Free_Kalibri [48]
4 years ago
5

Now the same particle is removed from the thread and placed over the center of a charged plate. Are there any conditions under w

hich it is possible for the particle to be suspended in the air above the plate? Show any relevant calculations and explain your reasoning.
Physics
1 answer:
labwork [276]4 years ago
6 0

Answer:

changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

Explanation:

When the particle is removed from the wire, friction can be electrically charged, either with negative charges (extra electrons) or with positive charge by electron removal, in this case when the particle is between the condenser plates it experiences a force due to the electric field given by

          ΔV = E d

Where ΔV is the potential difference, d the distance between the plates and E the electric field.

In these cases we can use Newton's second law, where the acceleration is zero

                F_{e} –W + B = 0

                 F_{e} = W –B

                 q E = mg - ρ_air g V

dodne B is the hydrostatic thrust

if  we know the density of the particular

                   ρ_particle = m / V

                   m = ρ_particle V

We replace

                 q E = g v (ρ_particle - ρ_air)

Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

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Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.6 cm and a current of 12 A. The bigger lo
puteri [66]

Answer:

the radius of bigger loop = 6 cm

Explanation:

given,

two concentric current loops

smaller loop radius = 3.6 cm

]current in smaller loop = 12 A

current in the bigger loop = 20 A

magnetic field at the center of loop = 0

Radius of the bigger loop = ?

B_t = B_1 + B_2

0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}

now, on solving

\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}

R_2 = I_2\dfrac{R_1}{I_1}

       = 20\times \dfrac{3.6}{12}

       = 6 cm

hence, the radius of bigger loop = 6 cm

7 0
3 years ago
A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2
SVEN [57.7K]

Answer:

K=512J

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass m_1 has an initial velocity v_{1i} and a final velocity v_{1f} and the block of mass m_2 has an initial velocity v_{2i} and a final velocity v_{2f} then the initial and final momentum of the system will be:

p_i=m_1v_{1i}+m_2v_{2i}

p_f=m_1v_{1f}+m_2v_{2f}

Since momentum is conserved, p_i=p_f, which means:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We know that the block is brought to rest by the collision, which means v_{2f}=0m/s and leaves us with:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}

which is the same as:

v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s

So kinetic energy of the bullet as it emerges from the block will be:

K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J

6 0
4 years ago
how many bits are required to sample an incoming signal 4000 times per second using 64 different amplitude level
Gelneren [198K]

Answer:

6 bits

Explanation:

The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.

Also, Sampling frequently (also known as oversampling) can help in improving signal quality.

To get the number of bits, we use:

2ⁿ = amplitude level

where n is the number of bits.

Given an amplitude level of 64, hence:

2ⁿ = 64

2ⁿ = 2⁶

n = 6 bits

6 0
3 years ago
A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
ddd [48]

It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a)the angular velocity, ω = 2600 rpm = \frac{2600 * 2\pi }{60} rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b)The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c)The angular acceleration, a=\frac{v^{2} }{r}

a = \frac{(47.62)^{2} }{0.175}

⇒a = 12958.08 m/s²

Learn more about angular velocity here:

brainly.com/question/13649539

#SPJ1

6 0
2 years ago
4.) A hydroelectric dam runs water thru turbines that are connected to
omeli [17]

Answer:

Explanation:

A turbine and generator produce the electricity

"A hydraulic turbine converts the energy of flowing water into mechanical energy. A hydroelectric generator converts this mechanical energy into electricity.

8 0
3 years ago
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