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2 years ago

7

. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north.

How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position?

1 answer:

KiRa [710]2 years ago

7 0

Answer:

H = 30.8 m

$\theta =54.25^0$

Explanation:

given,

first walks straight west = 18 m

then straight in north = 25 m

the distance from the start point

H² = B² + P²

H² = 18² + 25²

$H = \sqrt{18^2+25^2}$

$H = \sqrt{324+625}$

H = 30.8 m

the angle between

$tan \theta = \dfrac{P}{B}$

$tan \theta = \dfrac{25}{18}$

$\theta = tan^{-1}(\dfrac{25}{18})$

$\theta =54.25^0$

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Answer:

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The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

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2 years ago

A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.

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Hope it helps you

Good luck

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2 years ago

An object weighs 79.1 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 21.8 N.

Free_Kalibri [48]

Answer:

(a). The density of the object is 1382 kg/m³.

(b). The density of the oil is 536.4 kg/m³.

Explanation:

Given that,

Weight in air = 79.1 N

Weight in water = 21.8 N

Weight in oil = 48.4 N

We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{water}$

$F_{b}=79.1-21.8$

$F_{b}=57.3\ N$

$F_{b}=\rho g h$

Put the value into the formula

$57.3=1000\times V\times 9.8$

$V=\dfrac{57.3}{1000\times9.8}$

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We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho Vg$

$79.1=\rho\times5.84\times10^{-3}\times9.8$

$\rho=\dfrac{79.1}{5.84\times10^{-3}\times9.8}$

$\rho=1382\ kg/m^3$

The density of the object is 1382 kg/m³.

(b). We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{oil}$

$F_{b}=79.1-48.4$

$F_{b}=30.7\ N$

We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho_{oil} Vg$

$30.7=\rho_{oil}\times5.84\times10^{-3}\times9.8$

$\rho_{oil}=\dfrac{30.7}{5.84\times10^{-3}\times9.8}$

$\rho=536.4\ kg/m^3$

The density of the oil is 536.4 kg/m³.

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