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2 years ago

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In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

rts an upward force of 400 N. (a) What is the mass of the car in kilograms? (b) What is its weight in pounds?

1 answer:

Nata [24]2 years ago

6 0

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

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Answer:

7 deg

Explanation:

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$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

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To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

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Where,

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$m = 5.972*10^{24}kg$ Mass of Earth

$h = 56*10^6m$ Height

$r = 6.378*10^6m$ Radius of Earth

From the conservation of energy:

$\frac{1}{2} mv^2 = \frac{GMm}{d}$

Rearranging to find the velocity,

$v = \sqrt{\frac{2Gm}{d}} \rightarrow$ Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

$d = r+h$

$v = \sqrt{\frac{2Gm}{r+h}}$

Replacing the values we have that

$v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}$

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2 years ago

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Answer:

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Explanation:

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Given;

r = 6/2 = 3 cm = 0.03 m

h = 11 cm = 0.11 m

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2 years ago

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Answer:

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Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
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⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

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⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

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2 years ago