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3 years ago

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

1 answer:

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

ra1l [238]

The most exact answer is 78.4J also in this kind of options we can say answer "d"

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3 years ago

Which of these is not a part of the<br> cardiovascular system?

lilavasa [31]

Answer:I don't see any

Explanation:

6 0

3 years ago

Read 2 more answers

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia

Gala2k [10]

Answer:

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

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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