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3 years ago

Which data can be measured quantitatively

Physics

2 answers:

marissa [1.9K]3 years ago

3 0

all of it will aload

nadya68 [22]3 years ago

3 0

Answer:

the answer is above mine, hope this helped! :)

Explanation:

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?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

How is the mass number calculated for an element?

natali 33 [55]

# of protons + # of neutrons = atomic mass

8 0

4 years ago

Please explain your awnser:) Thanks in advance!

podryga [215]

The skateboarder will have the greatest potential energy at point G.

Potential energy is the energy possessed by a body by virtue of its position. The gravitational potential energy of a body is given by the expression,

$E_P=mgh$

Here, m is the mass of the body, g the acceleration due to gravity and h is the height of the body from the ground.Thus as h increases, the potential energy increases.

From the figure, it can be seen that the highest point in the skateboarder's trajectory is G. hence, he would have the highest potential energy at the point G.

4 0

4 years ago

A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the

MArishka [77]

Answer:

The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is

$I_c = \frac{I}{I_o} =8.48 *10^{-8}$