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Ierofanga [76]
3 years ago
8

Isaac Newton published his work on gravity in the Philosophiæ Naturalis Principia Mathematica (Latin for Mathematical Principles

of Natural Philosophy) in July 1687. In 2-3 complete sentences, detail observations you can make today that would lead you to the same conclusions Newton came to about gravity.
Physics
1 answer:
Igoryamba3 years ago
6 0
Today we can make observations that would lead us to the same conclusion as newton. If you throw a baseball in the air, it always comes back down. This is because of gravity. duh.
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A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

\theta=274rev

Explanation:

From the question we are told that:

Angular velocity \omega=510rpm

Mass m=40.kg

Diameter d 75=>0.75m

Off Time t=40.0s

Oscillation at Power off N=210

Generally the equation for Angular displacement is mathematically given by

 \theta_{\infty}=\frac{w+w_0}{t}t

 w=\frac{2*\theta_{\infty}}{t}-w_0

 w=\frac{28210}{40*(\frac{1}{60})}-510

 w=120rpm

Generally the equation for Time to come to rest is mathematically given by

 t=(\frac{\omega_0}{\omega_0-\omega})t

 t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})

 t=0.87min

Therefore Angular displacement is

 \theta =(\frac{120+510}{2})0.87

 \theta=274rev

6 0
3 years ago
i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it
Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

6 0
3 years ago
A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

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hope it helps

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