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3 years ago

9

a toy car moves around a loop-the-loop track.The loop is 0.5 high.What is the minimum speed of the car at the top of the loop fo

r it to stay on track?

1 answer:

Sunny_sXe [5.5K]3 years ago

3 0

Answer:

The speed of the car at the apex of the loop must be grater than 2.45 m/s

Explanation:

In order for the car to not fall off the track at the apex of the loop, the norm force of the track at the apex must be greater than zero.

Assuming frictionless life on the track which is also to have a perfectly circular shape near the top (radius being 0.25m), the norm force of the track and gravity both point down and result in the centripetal force:

$F_c=F_N+F_g$

The formula for centripetal force on a circular trajectory is

$F_c = m\frac{v^2}{r}$

and so the condition for the car to stay on the track can be written as

$m\frac{v^2}{r} = F_N + mg\implies F_N = m\frac{v^2}{r}-mg>0\\\implies |v| >\sqrt{gr}=\sqrt{9.8\frac{m}{s^2}0.25m}=2.45\frac{m}{s}$

The speed of the car at the apex of the loop must be grater than 2.45 m/s

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At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assum

Verdich [7]

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

$W_c=Mgh$ (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

$W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J$ (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

$0.16(E_c)=4,164,294.4J$

Ec: Calories

You solve for Ec and convert the result to Cal:

$E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal$

The amount of Calories needed by the climber was 6220.56 kcal

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2 years ago

How do you increase the potential energy of an apple

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To increase the potential energy of any object, not limited to an apple, you have to hold it higher.

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3 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

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For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

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Answer:

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2 years ago

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Answer:

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Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

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