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saveliy_v [14]
4 years ago
6

A small piece of dust of mass m = 4.1 µg travels through an electric air cleaner in which the electric field is 466 N/C. The ele

ctric force on the dust particle is equal to the weight of the particle.
(a) What is the charge on the dust particle?
C

(b) If this charge is provided by an excess of electrons, how many electrons does that correspond to?
electrons
Physics
1 answer:
pantera1 [17]4 years ago
7 0

Answer:

Explanation:

mass, m = 4.1 x 10^-6 g

Electric field, E = 466 N/C

Electric force = weight

(a) Electric force = q x E = m x g

where, q be the charge on the dust particle.

q x 466 = 4.1 x 10^-6 x 9.8

q = 8.62 x 10^-8 C

(b) Number of electron  = charge / charge of one electron

n = 8.62 x 10^-8 / (1.6 x 10^-19)

n = 5.4 x 10^11

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A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra
son4ous [18]

Answer:

The answer is v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}.

Explanation:

From the law of gravity,

F = G \frac{Mm}{r^2}

considering F as a conservative force, F = - \nabla U,

the general expression for gravitational potential energy is

U = -G \frac{Mm}{r},

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

g = -G \frac{M}{r^2},

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U = g M_e R,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

U_p = g M_e(R- r_s),

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8 0
3 years ago
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

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Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

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mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

5 0
3 years ago
Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0
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The energy of a light wave is calculated using the formula
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For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
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choli [55]

Answer:

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4 years ago
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