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3 years ago

suppose you increase your walking speed from 5m/s to 14m/s in a period of 1 . what is your acceleration?

1 answer:

7 0

Acceleration = (change in speed) / (time for the change.

change in speed = (ending speed) - (starting speed) = 9 m/s.

Acceleration = (9 m/s) / (period of 1) .

We don't know the units of the 'period of 1'.
If it means '1 second', then the acceleration is 9 m/s² .

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3 years ago

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Worth 50 point!!!

stira [4]

Answer:

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3 years ago

a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector

Delicious77 [7]

Answer:

magnitude=34.45 m

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Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance $d$ between two points:

$d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}$ (1)

$d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}$ (2)

$d=\sqrt{1186.81 m^{2}}$ (3)

$d=34.45 m$ (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

$tan \theta=\frac{Y2-Y1}{X2-X1}$ (5)

$tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}$ (6)

$tan \theta=\frac{24.8}{19.5}$ (7)

Finding $\theta$ :

$\theta= tan^{-1}(\frac{24.8}{19.5})$ (8)

$\theta= 55.52\°$ (9) This is the direction of the vector

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Explanation:

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