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3 years ago

Which term describes information recorded during an experiment?

Physics

2 answers:

Illusion [34]3 years ago

7 0

The answer is A, data!

Mrac [35]3 years ago

4 0

Data !
hope this helped <3

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ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo

vaieri [72.5K]

Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement = 2m + (-3)m. Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement is 1m behind his original starting point.

4 0

3 years ago

Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers

bekas [8.4K]

Answer: The correct answer is Option b.

Explanation:

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight

l = length of wire

A = area of cross section

$\Delta l$ = change in length

Hence, the correct answer is Option b.

6 0

3 years ago

A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m

anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

F = mv²/r

Where v²/r is the acceleration component of the force

a = v²/r

Substituting the values in the above equation

13.33 = v²/30

v² = 13.33 x 30

v² = 399.9

v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago

Which of the following wire gage sizes is the thickest? A. 8 B. 10 C. 14 D. 0

maks197457 [2]

A because it's the smaller the thicker you just can't have 0 gage

5 0

3 years ago