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3 years ago

6

a roller coaster car rapidly picks up speed as it rolls down a slope. and it starts down the slope, its speed is 4 m/s. but 3 se

conds later, at the bottom of the slope, its speed is 22m/s what is the acceleration?

1 answer:

slamgirl [31]3 years ago

3 0

I hope this helps.

a=(V2-V1)/t

a=(22-4)/3

A=18/3

A=6m/s/s

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An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

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Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

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When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

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(Rounded to $\text{$3$ sig. fig.}$ )

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Tju [1.3M]

m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

F = normal force by the surface on the person

f = kinetic frictional force acting on the person by the surface

μ = Coefficient of kinetic friction = 0.45

The normal force by the surface in upward direction balances the weight of the person in down direction , hence

F = mg eq-1

kinetic frictional force on the person acting is given as

f = μ F

using eq-1

f = μ mg

inserting the values

f = (0.45) (82) (9.8)

f = 361.6 N

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