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2 years ago

13

Identify which method of thermal energy transfer would be fastest through a vacuum, which would be fastest through a gas, and wh

ich would be fastest through a solid.

1 answer:

Blizzard [7]2 years ago

6 0

Answer:

vacuum-radiation

gas-convection

solid-conduction

Explanation:

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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.2 µC. They are separated by a distance of 0.28 m, and partic

NARA [144]

Answer:

q2 = - 8 × $10^{-6}$ C

negative sign because attract together

Explanation:

given data

q1 = +3.2 µC = 3.2 × $10^{-6}$ C

distance r = 0.28 m

force F = 2.9 N

to find out

q2 (magnitude and sign)

solution

we know that here if 2 charge is unlike charge

than there will be electrostatic force of attraction , between them

now we apply coulomb law that is

F = $\frac{q1q2}{4\pi \epsilon *r^{2}}$ ............1

here we know $\frac{1}{4\pi \epsilon}}$ = 9 × $10^{9}$ Nm²/C²

so from equation 1

2.9 = 9 × $10^{9}$ × $\frac{3.2*10^{-6}*q2}{0.28^{2}}$

q2 = $\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}$

q2 = - 8 × $10^{-6}$ C

4 0

3 years ago

The height of a cone is increasing at a rate of 10 cm/sec and its radius is decreasing so that its volume remains constant. How

ki77a [65]

Answer:

dr/dt = -2 cm/s.

Explanation:

The volume of a cone is given by:

$V=\frac{1}{3} \pi r^{2}h$ (1)

r is the radius
h is the height

Let's take the derivative with respect to time in each side of (1).

$\frac{dV}{dt}=\frac{1}{3} \pi \frac{d}{dt}(r^{2}h)=\frac{1}{3} \pi \left(2r\frac{dr}{dt}h+r^{2}\frac{dh}{dt} \right)$ (2)

We know that:

dh/dt = 10 cm / s (rate increasing of height)
dV/dt = 0 (constant volume means no variation with respect of time)
r = 4 cm
h = 10 cm

We can calculate how fast is the radius changing using the above information.

$0=\frac{1}{3} \pi \left( 2\cdot 4\cdot \frac{dr}{dt} \cdot 10 + 4^{2}\cdot 10)\right$

Therefore dr/dt will be:

$\frac{dr}{dt}=-\frac{160}{80}=-2 cm/s$

The minus signs means that r is decreasing.

I hope it helps you!

6 0

3 years ago

A particle of mass 2kg resting on a smooth table attached to a fixed point on the table by a rope 1.0m making 300revolution per

zepelin [54]

Answer is: 1973.17N aprox.
step by step in the pic below

7 0

3 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Dovator [93]

It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>

3 0

3 years ago

Read 2 more answers

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch

IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0

3 years ago