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babymother [125]
2 years ago
5

You sit "at rest" in front of your computer to answer this question. But you sit on the surface of a planet that spins, so even

while you sit "at rest," you are really hurtling through space at high speed. The average radius of the Earth is 6378 km. Ignoring all rotations except the spin of the Earth, what is your linear speed, in kph (kilometers per hour), while you sit in your home?
Physics
1 answer:
igomit [66]2 years ago
4 0

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

So, we can get v, as follows:

v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

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Upward pull of 850 N on a 81.6 kg bale of hay. What is the magnitude of the bales acceleration?
Helen [10]

Weight = (mass) x (acceleration of gravity).

When I calculate the weight of the 81.6 kg, the number I use for gravity
is 9.807 m/s².  That gives a weight of 800.25 N, so I think that's where the
question got the crazy number of 81.6 kg ... whoever wrote the problem
wants the hay to weigh 800 N, and that's what I'll use for the weight.

The forces on the bale of hay are gravity: 800N downward, and the
guy on the truck with the pitchfork pulling upward on it with 850 N. 
The net force on the bale is (850 - 800) = 50 N upward.

Use Newton's second law of motion:  (Net force) = (mass) x (acceleration)

Divide each side by 'mass' :     
      
                        Acceleration = (net force)/(mass)

On the hay wagon,             

                        Acceleration = (50 N upward) / (81.6 kg) = <em>0.613 m/s² upward</em> 

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3 years ago
A small hole is made at the bottom of a plastic cup. If it is filled with water and allowed to fall freely, will waterfall down
Alex_Xolod [135]

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5 0
2 years ago
A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

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