A merry-go-round initially at rest at an amusement park begins to rotate at time t=0. The angle through which it rotates is desc

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

   This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E =

where v is the final velocity, is the initial velocity, and m is the mass.

<p→ v = 20 m/s and = 0 m/s.<p→ K.E =

→ K.E =

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

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The expected measurements should range as follows: 5, 10, 15, 20, and 25 meters.

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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Answer:

According to the guideline for kilometers, every three seconds between a lightning strike and the subsequent thunder indicates the distance to the flash in kilometers.

Explanation:

To calculate the speed of sound in meters per second, we need to utilize certain conversion factors. One mile corresponds to 5 seconds after witnessing the lightning. Furthermore, 1 mile comprises 5280 feet, and 1 foot is equivalent to 0.3048 meters. This information is sufficient to solve the issue. The conversion ratios can be set up like this:

Observe how the ratios are organized such that the units cancel out during calculations. One ratio has miles in the numerator while the other has them in the denominator, leading to cancellation. The same applies to the feet.

The question requires us to provide the answer to one significant figure, resulting in the speed of sound rounding to 300m/s.

For the second part, we will again utilize conversions. This time we will set our ratios in reverse and realize that there are 1000 meters in 1 kilometer, leading us to:

This signifies that for every 3.11 seconds, the distance to the lightning strike is 1 kilometer. Since this is a fabric of general knowledge, we round to the nearest whole number for simplicity, establishing the guideline:

According to the rule for kilometers, every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 .

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.