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2 years ago

5

Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

of work on a 0.0002-C charge.

1 answer:

kondaur [170]2 years ago

7 0

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{12}{0.0001}$

$V=12\times 10^4\ Volt$

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{24}{0.0002}$

$V=12\times 10^4\ Volt$

Therefore, this is the required solution.

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Correct answer choice is :

C) A jar with snails crawling on living plants

Explanation:

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2 years ago

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A 50.0 kg crate is being pulled along a horizontal, smooth surface. The pulling force is 10.0 N and is directed 20.0 degree abov

allsm [11]

Explanation:

It is given that,

Mass of the crate, m = 50 kg

Force acting on the crate, F = 10 N

Angle with horizontal, $\theta=20^{\circ}$

Let N is the normal force acting on the crate. Using the free body diagram of the crate. It is clear that,

$N=mg-F\ sin\theta$

$N=50\times 9.8-10\ sin(20)$

N = 486.57 N

or

N = 487 N

If a is the acceleration of the crate. The horizontal component of force is balanced by the applied forces as :

$ma=F\ cos\theta$

$a=\dfrac{F\ cos\theta}{m}$

$a=\dfrac{10\times \ cos(20)}{50}$

$a=0.1879\ m/s^2$

or

$a=0.188\ m/s^2$

So, the normal force the crate and the magnitude of the acceleration of the crate is 487 N and $0.188\ m/s^2$ respectively.

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2 years ago

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

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2 years ago

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

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<span>The angular momentum of a particle in orbit is

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Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

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Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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2 years ago

⦁ A speed skater increases her speed form 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radi

Thepotemich [5.8K]

Answer:

7.85 m/s^2

Explanation:

linear or tangential acceleration= dv/dt

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⇒ $a_r =\frac{12.5^2}{20}$

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total acceleration

$a_T= \sqrt{a_t^2+a_r^2}$

putting values we get

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