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ruslelena [56]
3 years ago
15

As an tempature decreases what happens to the rate of radiation?

Physics
1 answer:
Firlakuza [10]3 years ago
5 0
As the temperature decreases, the rate of radiation goes down, but the radiation exists as long as the temperature is above the absolute zero, which is actually 0 Kelvin. 0 Kelvin equals -273°C or -460°F. All objects in the world radiate if above that temperature.
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A 4-kg cat is resting on a bookshelf that is 2 m high. What is the cats gravitational potential energy relative to the floor if
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The most exact answer is 78.4J also in this kind of options we can say answer "d"
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Which of these is not a part of the<br> cardiovascular system?
lilavasa [31]

Answer:I don't see any

Explanation:

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3 years ago
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Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia
Gala2k [10]

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Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

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3 years ago
In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
2 years ago
The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
3 years ago
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