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n200080 [17]
3 years ago
11

Tyler is in downtown San Antonio entertaining the tourists with amazing feats of physics. He carries a nickel

Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

The velocity of the nickel just before it hits the ground is approximately 49.3 m/s

Explanation:

In order to calculate the velocity of the nickel just before it hits the ground, we recall the kinematic equation of motion, v² = u² + 2·g·h, where the variables of the equation are defined as follows;

v = The velocity of the nickel just before it hits the ground after it is pushed off the observation deck

u = The initial velocity of the nickel just before it is pushed off the observation deck = 0 m/s

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which he nickel is pushed off the edge = 124 m

Substituting the given values and the constant, "g", we have;

v² = 0² + 2 × 9.8 × 124 = 2,430.4

v = √2,430.4

Using a graphing calculator, we have;

v = √2,430.4 = 14·√(62/5) ≈ 49.3

The velocity of the nickel just before it hits the ground = v ≈ 49.3 m/s.

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Juli2301 [7.4K]

Explanation:

it's a unit used to measure charge (C)

1C=1000millicoulombs

1millicoulomb=1000microcoulumbs

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2 years ago
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A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0
3 years ago
A 7.0kg skydiver is descending with a constant velocity
Vikentia [17]

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude mg, in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, R, in the upward direction

So the net force on the skydiver is:

F=mg-R

where

m = 7.0 kg is the mass

g=9.8 m/s^2

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

F=ma

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

a=0

Therefore the net force is zero:

F=0

And so, we have:

mg-R=0

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

R=mg=(7.0)(9.8)=68.6 N

6 0
3 years ago
Which situation is contrary to Newton’s first law of motion?
Bond [772]

Answer:

An object at rest stays at rest as long as unbalanced forces act on it.

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's First Law of Motion is known as Law of Inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of a physical object such as a truck is greatly dependent or influenced by its mass; the higher the quantity of matter in a truck, the greater will be its tendency to continuously remain at rest.

Hence, the situation which is contrary to Newton’s first law of motion is that, an object at rest stays at rest as long as unbalanced forces act on it.

According to Newton’s first law of motion, an object at rest stays at rest as long as unbalanced forces do not act on it.

4 0
2 years ago
Linea de tiempo sobre la aportaciones griegas a la astronomia y astrologia
rusak2 [61]

Answer:

1. 100 CE

Menelaus of Alexandria lived. a Greek mathematician and astronomer

2. 190 BCE - 120 BCE

Hipparchus of Nicea, an Hellenic language mathematician, astronomer and geographer, regarded by many historians as a scientist of the most effective quality and one amongst the most effective astronomical genius amongst ancient Greeks.

3. 276 BCE - 195 BCE

Eratosthenes, an Hellenic language Alexandrian scholar, who was a native of Cyrene and one amongst the most effective geographers in antiquity.

4. c. 310 BCE - c. 230 BCE

Aristarchus of Samos. A Greek astronomer and mathematician

5. 384 BCE - 322 BCE

Aristotle Era.

6. c. 571 BCE - c. 497 BCE

Pythagoras of Samos lived during this era.

7. 585 BCE

Media and Lydia went into battle and broke off immediately as a result an entire eclipse of the sun which occurred causing the two armies to create peace. The eclipse was already predicted by Thales of Miletus.

8. 585 BCE

Thales of Miletus lived during now.

Explanation:

Ancient Greeks were some of the first people known to study the sky and understand what astronomy really entails. They discovered the Earth was spherical in shape and went ahead to devise a means to measure its size. They also were the ones who created the idea of a geocentric solar system, which was incorrect, But assisted us in understanding the universe for over hundreds of years.

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3 years ago
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