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Alexxx [7]
3 years ago
6

On the way home from school, Taylor's car runs out of gas. He has to walk 25m north and 10m west in order to reach the nearest g

as station. Find his distance traveled and his displacement from his car
Physics
1 answer:
spin [16.1K]3 years ago
8 0

Answer:

<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>

Explanation:

<u>Displacement  and Distance</u>

These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is calculated independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:

d = 25 m + 10 m = 35 m

To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.

The magnitude of the displacement is the distance from (0,0) to (-10,25):

D=\sqrt{(25-0)^2+(-10-0)^2}

D=\sqrt{625+100}=\sqrt{725}

D = 26.93 m

The distance is 35 m and the magnitude of the displacement is 26.93 m

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Answer:

Force = mass × acceleration

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{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

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A biconvex lens is formed by using a piece of plastic(n=1.70).
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Answer:

f =17.15\ cm

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

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focal length= ?

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})

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\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})

\dfrac{1}{f}=0.70 \times 0.0833

f = \dfrac{1}{0.7 \times 0.0833}

f =17.15\ cm

the focal length of the lens is equal to 17.15 cm

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f_d = 400(\frac{(343+18)}{(343-35)})

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