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miskamm [114]
2 years ago
6

A cross country skier moves 32 m north, then 65 m south, and finally 16 m north. What is the distance? What is the displacement

?
Physics
1 answer:
Fofino [41]2 years ago
4 0

The distance and Displacement travelled by skier is 17m and 16 m

<h3>What is Distance and Displacement?</h3>

Distance is the length of the actual path taken by object.

Displacement is the shortest distance between the initial and final position.

The SI unit of both distance and displacements is meter (m).

In this case,

Distance = 32m - 65m + 16 m

                 = 17 m

Here ,

Displacement  = Initial position - Final position

= 32m -16m  = 16m

Hence , The distance and displacement of the skier are 17m and 16m.

For more distance and displacement queries visit:

brainly.com/question/18965361

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A flask is filled with 1.57 L (L: liter) of a liquid at 90.6 °C. When the liquid is cooled to 10.2 °C, its volume is only 1.38 L
nydimaria [60]

The coefficient of volume expansion of the liquid = 1.97 × 10⁻³ /⁰C

<h3>What is the coefficient of volume expansion of the liquid?</h3>

The amount of volume that a substance expands by per unit of its original volume for each degree that its temperature rises is known as the coefficient of volume expansion.

As we know, The coefficient of volume expansion of the liquid

= change in volume/(original volume × temperature difference)

= (V₂ - V₁)/[V₁ × (T₂ - T₁)]

Change in volume = (V₂ - V₁)

Here, V₂ ( final volume) = 1.31 L

V₁  (initial volume)= 1.55 L

T₂ (final temperature) = 14.7 degrees

T₁ (initial temperature) = 96 degrees

The coefficient of volume expansion of the liquid

= (1.31 - 1.55) / [1.5 × (14.7- 96)]

= 1.97 × 10⁻³ /⁰C

Thus, the coefficient of volume expansion of the liquid = 1.97 × 10⁻³ /⁰C

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8 0
2 years ago
4. In 1994, element 111 was discovered by an international team of physicists. Its
otez555 [7]

The distance between the similar charges is 0.753 m. The electric force is the inverse of the square of the distance.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is;

q₁,q₂ are the charges = 1.6 ×10⁻¹⁹ C

R is the distance

F is the electric force=2.0×10²⁴ N

The electric force is found as;

\rm F = \frac{Kq_1q_2}{R^2} \\\\ \rm 2.0 \times  10^{28}= \frac{9\times 10^9 \times 1.6 \times 10^{-19}1.6\times 10^{-19} }{R^2} \\\\ R =0.753 \ m

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