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3 years ago

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A car is moving with a speed of 22 m/s. The driver then brakes, and the car comes to a halt after 6.5 s. What is the distance co

vered by the car after the driver brakes, until it comes to a stop? A) 36 m B) 72 m C) 140 m D) 210 m

1 answer:

Lena [83]3 years ago

8 0

The answer would be 72.

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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.

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(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

$57.3=1000\times V\times 9.8$

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

$V=5.84\times10^{-3}\ m^3$

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.

Hence, This is the required solution.

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