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3 years ago

The process of heat radiation is the ONLY method of heat transfer that can occur _____.

Physics

2 answers:

Neko [114]3 years ago

7 0

B) through space I am sorry if it is wrong

dimaraw [331]3 years ago

6 0

Answer:

B) through space.

Explanation:

By the process of radiation, heat moves through space in the form of electromagnetic waves. Electromagnetic waves, unlike compression waves, can move through empty space.

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Why can someone standing behind a mirror not see the image?

ANTONII [103]

Answer:

The image behind the mirror is called a virtual image because it cannot be projected onto a screen—the rays only appear to originate from a common point behind the mirror. If you walk behind the mirror, you cannot see the image, because the rays do not go there

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3 years ago

Characteristics of highly sensitive galvonometer

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A resistance of 990ohm is increased by 10

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3 years ago

Gravitational energy can be negative?

AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

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3 years ago

Particles of m1 and m2 (m2>m1) are connected by a line in extensible string passing over a smooth fixed pulley. Initially, bo

Tresset [83]

Answer:

The velocity with which the mass will hit the floor is $v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}$

Explanation:

If the tension in the string is $T$ , for $m_1$ we have

$T- m_1g =m_1a$ ,

and for the mass $m_2$

$T -m_2g = -m_2a$

From these equations we solve for $a$ and get:

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

The kinematic equation

$v_f^2 = v_0^2+2ax$

gives the final velocity $v_f$ of a particle, when its initial velocity was $v_0$ , and has traveled a distance $x$ while undergoing acceleration $a$ .

In our case

$v_0 = 0$ (the initial velocity of the particles is zero)

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

which gives us

$v_f^2 = 2ax$

$v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g$

$\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }$

which is the velocity with which the mass $m_2$ will hit the floor.

8 0

3 years ago

With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be

jeyben [28]

Question:

Assumptions

Voltage of battery = 24 V

Resistance on the right,20Ω parallel to 10 Ω resistor

Answer:

For the current out of the battery to be the same as when the switch was opened with the switch closed, the resistance on the resistor on the right must be approximately 20/3 Ω

Explanation:

We note that the switch in the assumption is on the same line as the 20 Ω resistor.

With a voltage of 24 V, and the switch closed, we have;

Total resistance = $\frac{1}{\frac{1}{10} +\frac{1}{20} } = \frac{20}{3} \Omega$

Current out of voltage, I = Voltage/(total resistance)

= 24 ÷ 20/3 = 24 × 3/20 = 18/5 A

Therefore, with the switch opened, we get

Resistance on the right = Initial total resistance $= \frac{20}{3} \Omega$

Therefore, with the switch opened, the resistance on the resistor on the right must be approximately equal to the resultant resistance of the two resistances in parallel.

6 0

3 years ago