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3 years ago

When used in physical therapy, a diathermy machine gives the sensation of "deep heat" in the tissue being treated.

Physics

1 answer:

rewona [7]3 years ago

6 0

Answer:

A

Explanation

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1 kg of aluminum requires 887 J (Joule) to raise its temperature by 1 K (Kelvin). The specific heat of aluminum is expressed as

zavuch27 [327]

Answer:

887 J

Explanation:

the specific heat of a material is defined as the amount of heat energy required to raise the temperature of the substance by 1 K of mass 1 kg. So, the specific heat of the aluminium, c = 887 J/kg K

mass of aluminium, m = 100 g = 0.1 kg

T1 = - 9° C

T2 = 1° C

ΔT = 1 + 9 = 10° C

The amount of heat required

H = m x c x ΔT

H = 0.1 x 887 x 10 = 887 J

3 0

3 years ago

What's the answer for this question will give BRAILIEST!!!!!

fredd [130]

The third one! :) Hope you live up to what ya promised man lol.

6 0

4 years ago

A rope of negligible mass is wrapped around a 225-kg solid cylinder of radius 0.400 m. the cylinder is suspended several meters

emmainna [20.7K]

Let me know if you have questions

5 0

3 years ago

1. (1 pt) Which of these is a physical property of wood? A. Wood can burn. B. Wood can rot. C. Wood does not rust. D. Wood is so

anygoal [31]

A physical property can be seen or a change in color.
The answer is B because you can see when wood is rotting.

3 0

4 years ago

?a wire is stretched 30% what is the percentage change in resistance

Marat540 [252]

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

$l_{o}$ = 1.3l, and $A_{o}$ = $\frac{A}{1.3}$

So that;

$R_{o}$ = (ρ $l_{o}$ ) ÷ $A_{o}$ = (ρ × 1.3l) ÷ ( $\frac{A}{1.3}$ )

= (1.3lρ) ÷ ( $\frac{A}{1.3}$ )

= $(1.3)^{2}$ × [(ρl) ÷ A]

= 1.69R (∵ R = (ρl) ÷ A)

$R_{o}$ = 1.69R

Where $R_{o}$ is the new resistance, $l_{o}$ is the new length, and $A_{o}$ is the new area after stretching the wire.

The change in resistance of the wire = $R_{o}$ - R

= 1.69R - 1R

= 0.69R

The percentage change in resistance = $\frac{0.69R}{R}$ × 100

= 0.69 × 100

= 69%

The percentage change in resistance of the wire is 69%.

3 0

3 years ago