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2 years ago

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A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

ion m, what is the electric field contributed by the polarization charges on the surface of the metal sphere? (Express your answer in vector form.)

1 answer:

liraira [26]2 years ago

6 0

Answer: E = $1.8 *10 ^{4} N$

Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below

$E =\frac{Kq}{r^{2} }$ r

where $k=9* 10^{9}N/m^{2}$ , $q=2 *10 ^{-8} c$ , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

$E=\frac{9 * 10^{9}* 2* 10^{-8} }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N$

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When objects are heated they tend to expand because

densk [106]

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When objects are heated, their molecules tend to vibrate fast. As they vibrate, the space between each atom increases. This keeps on happening, and the object expands until it has cooled down.
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5 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

2 years ago

Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay

Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

$w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s$

The maximum velocity is

$v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s$

8 0

2 years ago

Plants need to....................synthesize

koban [17]

Answer:

nitrogen

Explanation:

because I also had this in exam and I was correct

5 0

2 years ago

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

$T=2\pi \sqrt{\frac{m}{k} }$

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0

5 months ago