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3 years ago

What is the displacement of a NASCAR stock car after he completes the Indianapolis 500?

1 answer:

3 0

I'm such a devoted NASCAR fan that I don't even know if the starting line and finish line at Indy are the same line ... or maybe they're in different places, like the 100-meter sprint in track.

I have no idea, but I'm going to guess that the start and finish at Indy are the SAME line.

If that's true, then the displacement of a car that runs the whole 500 miles is very close to ZERO.

Displacement is the distance and direction between the place where the object starts out and the place where it ends up. The route it follows to get from the start to the finish is completely ignored, and doesn't matter.

(Do they the Indianapolis 500 in "stock" cars ? ?)

You might be interested in

Charles' law explains which of these phenomena?

Luden [163]

I think the correct answer would be that Charles' law explains why a balloon deflates when the air around it cools. Charles' law is a simplification of the ideal gas law. At constant pressure, volume and temperature have a direct relationship. Hope this helps.

5 0

3 years ago

Which statement is correct? Theories are accepted as true when a single experiment yields similar results to another one. When a

levacccp [35]

The third statement is correct.

3 0

2 years ago

Read 2 more answers

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

P = 0.816 Watt

7 0

3 years ago

A 1400 kg wrecking ball hangs from a 20-m-long cable. the ball is pulled back until the cable makes an angle of 30.0 ∘ with the

anastassius [24]

From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m.

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

5 0

3 years ago

Read 2 more answers

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago